Definite Integration

$ \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{d x}{\sec ^{2} x+\left(\tan ^{2024} x-1\right)\left(\sec ^{2} x-1\right)} $

$ \begin{array}{l}\int_{\frac{2 \pi}{10}}^{\frac{3 \pi}{0}} \frac{d x}{1+\tan ^{2} x+\tan ^{2024} x-\tan ^{2} x} \\\\ =\int_{\frac{2 \pi}{10}}^{\frac{3 \pi}{10}} \frac{d x}{1+\tan ^{2024} x} \\\\ I=\int_{\frac{2 \pi}{10}}^{\frac{3 \pi}{0}} \frac{\cos ^{2024} x}{\sin ^{2024} x+\cos ^{2024} x} d x \end{array} $

Applyh P_IV

$ I=\int_{\frac{\pi}{10}}^{\frac{3 \pi}{10}} \frac{\sin ^{2024} x}{\cos ^{2024} x+\sin ^{2024} x} d x $

On adding Eqs. (i) and (ii), we get

$ \begin{array}{l} 2 I=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} d x=(x)_{\pi / 5}^{3 \pi / 10}=\frac{3 \pi}{10}-\frac{2 \pi}{10}=\frac{\pi}{10} \\\\ I=\frac{\pi}{20} \end{array} $

Apply P - IV

$ \begin{aligned} I & =\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \frac{\cos (-5 x)}{1+e^{-5 x}} d x \\\\ & =\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \frac{e^{5 x} \cos 5 x}{e^{5 x}+1} d x \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{array}{l} 2 I=\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5 x\left(e^{5 x}+1\right)}{e^{5 x}+1} d x=\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \cos 5 x d x \\\\ =2 \int_{0}^{\frac{\pi}{15}} \cos 5 x d x \\\\ 2 I=2 \frac{1}{5}[\sin 5 x]_{0}^{\frac{\pi}{15}} \\\\ I=\frac{1}{5}\left[\sin \frac{\pi}{3}-0\right]=\frac{1}{5} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{10} \end{array} $

To solve the given integral problem, we can start by simplifying the expression inside the integral:

$ \frac{3}{25} \int_{0}^{25 \pi} \sqrt{\left|\cos x - \cos^3 x\right|} \, dx $

We can rewrite this expression as:

$ \frac{3}{25} \int_{0}^{25 \pi} \sqrt{|\cos x (1 - \cos^2 x)|} \, dx $

Recognizing that $1 - \cos^2 x = \sin^2 x$, this becomes:

$ \frac{3}{25} \int_{0}^{25 \pi} \sqrt{|\cos x| \sin^2 x} \, dx $

We further simplify this to:

$ \frac{3}{25} \int_{0}^{25 \pi} |\sin x| \sqrt{|\cos x|} \, dx $

Next, we decompose the integral over the period from $0$ to $25\pi$ into parts over smaller intervals where the function symmetry and periodicity can be utilized:

$ = \frac{3}{25} \left[13 \int_{0}^{\frac{\pi}{2}} \sin x \sqrt{\cos x} \, dx + 13 \int_{\frac{\pi}{2}}^{\pi} \sin x \sqrt{-\cos x} \, dx - 12 \int_{\pi}^{\frac{3\pi}{2}} \sin x \sqrt{-\cos x} \, dx - 12 \int_{\frac{3\pi}{2}}^{2\pi} \sin x \sqrt{\cos x} \, dx\right] $

For each sub-interval, we compute the integral:

For $ \int_{0}^{\frac{\pi}{2}}$, use the substitution $\cos x$ from 1 to 0.

For $ \int_{\frac{\pi}{2}}^{\pi}$, and similarly mapped intervals $\int_{\pi}^{\frac{3\pi}{2}}$, $\int_{\frac{3\pi}{2}}^{2\pi}$, follow similar transformations respecting their respective limits to evaluate each segment.

The integrals resolve as follows:

$ = \frac{3}{25} \left[-13 \left[\frac{2(\cos x)^{\frac{3}{2}}}{3} \right]_{0}^{\frac{\pi}{2}} + 13 \left[\frac{2(-\cos x)^{\frac{3}{2}}}{3}\right]_{\frac{\pi}{2}}^{\pi} - 12\left[\frac{2(-\cos x)^{\frac{3}{2}}}{3}\right]_{\pi}^{\frac{3\pi}{2}} + 12 \left[\frac{2}{3}(\cos x)^{\frac{3}{2}} \right]_{\frac{3\pi}{2}}^{2\pi}\right] $

Evaluating each integral gives:

$ = \frac{3}{25} \times \frac{1}{3}\left[-26(0-1) + 26(1-0) - 24(0-1) + 24(1-0)\right] $

This simplifies to:

$ = \frac{1}{25}[26 + 26 + 48] = \frac{1}{25} \times 100 = 4 $

Thus, the final calculated value of the integral is 4.

To analyze the integrals, we start by considering the expression $ I = \int_{-2\pi}^{2\pi} \sin^m x \cos^n x \, dx $.

Since the given condition is:

$ I = 4 \int_{0}^{\pi} \sin^m x \cos^n x \, dx $

We know that:

$ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx $

if $ f(x) $ is an even function. This suggests that $ \sin^m x \cos^n x $ might be even in this case. To satisfy the condition where the integral over the larger symmetric interval is four times the integral over the smaller positive interval, $ m $ must be even.

Similarly, consider:

$ \int_{-\pi}^{\pi} \sin^r x \cos^s x \, dx = 4 \int_{0}^{\pi/2} \sin^r x \cos^s x \, dx $

Here, for a comparable reason, we must have $ r $ as even.

Finally, for:

$ \int_{-\pi/2}^{\pi/2} \sin^l x \cos^m x \, dx = 0 $

the function should be odd because the integral of an odd function over a symmetric interval around zero results in zero. Thus, $ l $ must be odd.

With the constraints $ 9 > m > l > s > n > r > 2 $, and that $ m $ and $ r $ are even while $ l $ is odd, a possible assignment of these integers could be:

$ m = 8 $

$ l = 7 $

$ s = 6 $

$ n = 5 $

$ r = 4 $

Next, let's test the option that matches this setup:

For option C, $(s-2)(s+2) = ln - 3$:

$ (s-2)(s+2) = (6-2)(6+2) = 4 \times 8 = 32 $

The expression $ ln - 3 $ was a typo originally intended to test our guess, but $ 32 = 32 $ checks out, confirming this is our desired setup.

Given, $ I=\int_{0}^{\pi}\left(\sin ^{3} x+\cos ^{2} x\right)^{2} d x $

$ \begin{array}{c} I=\int_{0}^{\pi}\left[\sin ^{3}(\pi-x)+\cos ^{2}(\pi-x)\right]^{2} d x \\\\ I=\int_{0}^{\pi}\left(\sin ^{3} x+\cos ^{2} x\right)^{2} d x \\\\ \Rightarrow I=2 \int_{0}^{\pi / 2}\left(\sin ^{3} x+\cos ^{2} x\right)^{2} d x \end{array} $

By applying the property,

$ I=2 \int_{0}^{\pi / 2}\left(\cos ^{3} x+\sin ^{2} x\right)^{2} d x $

On adding Eqs. from (i) and (ii), we get

$ \begin{array}{c} 2 I=2\left[\int _ { 0 } ^ { \pi / 2 } \left(\left(\sin ^{3} x+\cos ^{2} x\right)^{2}\right.\right. \\\\ \left.+\left(\cos ^{3} x+\sin ^{2} x\right)^{2}\right] d x \\\\ I=\int_{0}^{\pi / 2} \sin ^{6} x+\cos ^{6} x+\sin ^{4} x+\cos ^{4} x \\\\ +2 \sin ^{3} x \cos ^{2} x+2 \cos ^{3} x \sin ^{2} x d x \\\\ I=\int_{0}^{\pi / 2}\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x\right. \\\\ \left.+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right) \\\\ +\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} \end{array} $

$ x(\sin x+\cos x) d x $

$ I=\int_{0}^{\pi / 2} 2\left(\sin ^{4} x+\cos ^{4} x\right)+\sin ^{2} x \cos ^{2} $

$ x\left(2 \sin x+2 \cos x^{-1}\right) $

$ I=\int_{0}^{\pi / 2} 2\left(1-2 \sin ^{2} x \cos ^{2} x\right)+\sin ^{2} x \cos ^{2} $

$ x(2 \sin x+2 \cos x-1) d x $

$ \begin{array}{r} I=\int_{0}^{\pi / 2} 2-5 \sin ^{2} x \cos ^{2} x+2 \sin ^{3} x \cos ^{2} \\\\ x+2 \sin ^{2} x \cos ^{3} x d x \end{array} $

Let $ I=\int_{0}^{\pi / 2} 5 \sin ^{2} x \cos ^{2} x d x $

$ =\frac{5}{4} \int_{0}^{\pi / 2} \sin ^{2} 2 x d x $

$ =\frac{5}{4} \int_{0}^{\pi / 2} \frac{1-\cos 4 x}{2} d x $

$ =\frac{5}{8}\left(x-\frac{\sin 4 x}{4}\right)_{0}^{\pi / 2}=\frac{5}{8}\left[\frac{\pi}{2}\right]=\frac{5 \pi}{16} $

$ I_{2}=\int_{0}^{\pi / 2} 2 \sin ^{3} x \cos ^{2} x d x $

$ =\int_{0}^{\pi / 2}\left(1-\cos ^{2} x\right) \cos ^{2} x \sin x d x $

Let $ \cos x=t $

$ -\sin x d x=d t \Rightarrow \sin x d x=-d t $

Also, $ x=0, t=1 $

$ \begin{array}{l} x=\pi / 2, t=0 \\\\ \begin{aligned} I_{2} & =2 \int_{1}^{0}\left(t^{2}-1\right) t^{2} d t=2 \int_{1}^{0}\left(t^{4}-t^{2}\right) d t \\\\ & =2\left[\frac{t^{5}}{5}-\frac{t^{3}}{3}\right]_{1}^{0}=2\left[-\frac{1}{5}+\frac{1}{3}\right]=\frac{4}{15} \\\\ I_{3} & =2 I=\int_{0}^{\pi / 2} \sin ^{2} x \cos ^{3} x d x \\\\ I_{3} & =2 \int_{0}^{\pi / 2} \sin ^{2} x\left(1-\sin ^{2} x\right) \cos x d x \end{aligned} \end{array} $

Let $ \sin x=u \Rightarrow \cos x d x=d u $

Also, $ x=0, u=0 \Rightarrow x=\pi / 2, u=1 $

$ \begin{aligned} I_{3} & =2 \int_{0}^{1} u^{2}\left(1-u^{2}\right) d u \\\\ & =2\left[\frac{u^{3}}{3}-\frac{u^{5}}{5}\right]_{0}^{1}=2\left[\frac{1}{3}-\frac{1}{5}\right]=\frac{4}{15} \\\\ I & =[2 x]_{0}^{\pi / 2}-\frac{5 \pi}{16}+\frac{4}{15}+\frac{4}{15} \\\\ & =\pi-\frac{5 \pi}{16}+\frac{8}{15}=\frac{11 \pi}{16}+\frac{8}{15} \end{aligned} $

Given, $ I=\int_{-\frac{\pi}{8}}^{\pi / 8} \frac{\sin ^{4}(4 x)}{1+e^{4 x}} d x $

$ \begin{array}{l} I=\int_{0}^{\pi / 8}\left[\frac{\sin ^{4}(4 x)}{1+e^{4 x}}+\frac{\sin ^{4}(-4 x)}{1+e^{-4 x}}\right] d x \\\\ I=\int_{0}^{\pi / 8} \sin ^{4}(4 x)\left[\frac{1}{1+e^{4 x}}+\frac{e^{4 x}}{e^{4 x}+1}\right] d x \\\\ I=\int_{0}^{\pi / 8} \sin ^{4} 4 x d x \end{array} $

Let $ 4 x=t \Rightarrow 4 d x=d t $

If $ x=0 \Rightarrow t=0 $

$ \begin{array}{c} x=\frac{\pi}{8} \Rightarrow t=\frac{\pi}{2} \\\\ I=\frac{1}{4} \int_{0}^{\pi / 2} \sin ^{4} t d t \\\\ I=\frac{1}{4} \int_{0}^{\pi / 2} \sin ^{4}\left(\frac{\pi}{2}-t\right) \\\\ =\frac{1}{4} \int_{0}^{\pi / 2} \cos ^{4} t d t \end{array} $

$ \begin{array}{l} \quad I=\frac{1}{4} \int_{0}^{\pi / 2} \sin ^{4}\left(\frac{\pi}{2}-t\right) d t \\\\ \quad=\frac{1}{4} \int_{0}^{\pi / 2} \cos ^{4} t d t \end{array} $

On adding in Eqs (i) and (ii), we get.

$ \begin{array}{l} 2 I=\frac{1}{4} \int_{0}^{\pi / 2}\left(\sin ^{4} t+\cos ^{4} t\right) d t \\\\ I=\frac{1}{8} \int_{0}^{\pi / 2}\left[1-\frac{1}{2}(\sin 2 t)^{2}\right] d t \\\\ I=\frac{1}{8} \int_{0}^{\pi / 2}\left[1-\frac{1}{4}(1-\cos 4 t)\right] d t \\\\ I=\frac{1}{8} \int_{0}^{\pi / 2}\left[\frac{3}{4}+\frac{1}{4} \cos 4 t\right] d t \\\\ I=\frac{1}{8}\left(\frac{3 t}{4}+\frac{\sin 4 t}{16}\right)_{0}^{\pi / 2} \Rightarrow I=\frac{1}{8}\left(\frac{3 \pi}{8}\right)=\frac{3 \pi}{64} \end{array} $

Let $I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) d x \quad \ldots \text { (i) } $

$ \begin{aligned} & \Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log \left(\sin \left(4\left(\frac{\pi-6}{8}+\left(-\frac{3}{4}\right)-x\right)+3\right)\right) d x \\\\ & \Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log \left(\sin \left(\frac{\pi}{2}-(4 x+3)\right)\right) d x \\\\ & \quad \Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\cos (4 x+3)) d x \quad \ldots \text { (ii) } \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 2 I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3) \cos (4 x+3)) d x \\\\ & \Rightarrow 2 I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log \left(\frac{\sin (2(4 x+3))}{2}\right) d x \end{aligned} $

$ \begin{aligned} & \Rightarrow 2 I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\sin (2(4 x+3))) d x \\\\ & \qquad-\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}}(\log 2) d x \\\\ & \Rightarrow 2 I=I-(\log 2)\left[\frac{\pi-6}{8}-\left(-\frac{3}{4}\right)\right] \\\\ & \Rightarrow I=-\frac{\pi}{8} \log 2 \end{aligned} $

$ \begin{aligned} & \text {Let } I=\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x \\\\ & \text { Put } 1+\sqrt{x}=t \\\\ & \Rightarrow \quad \frac{1}{2 \sqrt{x}} d x=d t \\\\ & \Rightarrow \quad d x=2(t-1) d t \\\\ & \text { When } x=0, t=1 \\\\ & \text { When } x=16, t=5 \\\\ & \begin{aligned} \therefore I & =\int_1^{5 t-1} \\\\ t & (2(t-1)) d t \\\\ & =2 \int_1^{5 t^2-2 t+1} \\\\ & =2 \int_1^5\left(t-2+\frac{1}{t}\right) d t \\\\ & =2\left[\frac{t^2}{2}-2 t+\log t\right]_1^5 \end{aligned} \end{aligned} $

$ \begin{aligned} & =2\left[\left\{\frac{25}{2}-10+\log 5\right\}-\left\{\frac{1}{2}-2+\log 1\right\}\right] \\\\ & =2\left[\frac{5}{2}+\log 5+\frac{3}{2}+0\right] \\\\ & =8+2 \log 5 \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^{32 \pi} \sqrt{1-\cos 4 x} d x \\\\ & =\int_0^{32 \pi} \sqrt{1-\left(1-2 \sin ^2 2 x\right)} d x \\\\ & =\int_0^{32 \pi} \sqrt{2}|\sin 2 x| d x \end{aligned} $

The period of $|\sin 2 x|$ is $\frac{\pi}{2}$

$ \begin{aligned} \therefore \quad I & =64 \int_0^{\pi / 2} \sqrt{2} \sin 2 x d x \\\\ & =64 \sqrt{2}\left[\frac{-\cos 2 x}{2}\right]_0^{\pi / 2} \\\\ & =32 \sqrt{2}[-\cos \pi+\cos 0] \\\\ & =32 \sqrt{2}[-(-1)+1]=64 \sqrt{2} \end{aligned} $

We have,

$ \begin{aligned} f(x)= & \int \frac{\sin 2 x+2 \cos x}{4 \sin ^2 x+5 \sin x+1} d x \\\\ & =\int \frac{2 \cos x(\sin x+1)}{(4 \sin x+1)(\sin x+1)} d x \\\\ & =\int \frac{2 \cos x}{4 \sin x+1} d x \end{aligned} $

$\begin{aligned} & \text { Let } 4 \sin x+1=t \Rightarrow 4 \cos x d x=d t \\\\ & \begin{aligned} \Rightarrow & f(x)=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log t+C \\\\ & =\frac{1}{2} \log (4 \sin x+1)+C\end{aligned} \\\\ & \begin{aligned} \because & f(0)=0 \Rightarrow C=0 \\\\ \therefore & f(x)=\frac{1}{2} \log (4 \sin x+1)\end{aligned} \\\\ & f\left(\frac{\pi}{6}\right)=\frac{1}{2} \log \left(\frac{4}{2}+1\right)=\frac{1}{2} \log 3\end{aligned}$

Let $I=\int_{-2}^2 x^4\left(4-x^2\right)^{7 / 2} d x$

$ \begin{aligned} & I=2 \int_0^2 x^4\left(4-x^2\right)^{7 / 2} d x \\\\ & \qquad \quad[\because f(x) \text { is even function }] \\\\ & \text { put } x=2 \sin \theta \Rightarrow d x=2 \cos \theta d \theta \\\\ & \text { when } x=0, \theta=0 \text { and } x=2, \theta=\frac{\pi}{2} \\\\ & \therefore I=2 \int_0^{\pi / 2} 2^4 \sin ^4 \theta(2 \cos \theta)^7 \cdot 2 \cos \theta d \theta \\\\ & I=2 \cdot 2^4 \cdot 2^8 \int_0^{\pi / 2} \sin ^4 \theta \cos ^8 \theta d \theta \\\\ & I=2^{13} . \end{aligned} $

$ \frac{(4-1)(4-3)(8-1)(8-3)(8-5)(8-7)}{12 \times 10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2} $

(by Wall's formulae)

$ \begin{aligned} & =\frac{2^{13} \times 3 \times 1 \times 7 \times 5 \times 3 \times 1 \times \frac{\pi}{2}}{12 \times 10 \times 8 \times 6 \times 4 \times 2} \\\\ & =\frac{2^6 \times 3 \times 7 \times 5 \times 3 \times \pi}{6 \times 5 \times 4 \times 3 \times 2}=28 \pi \end{aligned} $

Let $I=\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$

As $\sin ^4 x+\cos ^4 x$ is periodic with $\frac{\pi}{2}$

So, $\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$

$ \begin{aligned} & =\int_0^{4 \times \frac{\pi}{2}}\left(\sin ^4 x+\cos ^4 x\right) d x \\ = & 4 \int_0^{\pi / 2}\left(\sin ^4 a x+\cos ^4 x\right) d x \\ = & 4 \int_0^{\pi / 2} 1-\frac{\sin ^2 2 x}{2} d x \\ = & 4 \int_0^{\pi / 2} \frac{4-1+\cos 4 x}{4} d x \\ = & 4 \int_0^{\pi / 2} \frac{3+\cos 4 x}{4} d x \\ = & {\left[3-\frac{\sin 4 x}{4}\right]_0^{\frac{\pi}{2}} } \\ = & 3\left[\frac{\pi}{2}\right]=\frac{3 \pi}{2} \end{aligned} $

Now, $\int_0^\pi \sin ^2 x d x=2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x$

$ \begin{aligned} & =2 \int_0^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) d x=\left[x-\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}} \\ & =\frac{\pi}{2} \end{aligned} $

Similarly,

$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \cos ^2 x d x=\frac{\pi}{4} \\ \therefore & \frac{3 \pi}{2}=K \times \frac{\pi}{2}+L \times \frac{\pi}{4} \\ & 2 K+L=6 \end{aligned} $

If $K=1$, then $L=4$ and if $K=2$, then $L=2$

$\therefore$ There are two possible ordered pair i.e. $(1,4),(2,2)$.

$ \text { } \begin{aligned}Let \,\, I & =\int_0^\pi \frac{x \sin x d x}{4 \cos ^2 x+3 \sin ^2 x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & =\int_0^\pi \frac{x \sin x d x}{4 \cos ^2 x+3-3 \cos ^2 x} \\ & =\int_0^\pi \frac{x \sin x}{\cos ^2 x+3} \\ & =\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{\cos ^2(\pi-x)+3} \end{aligned} $

$ \begin{aligned} &=\int_0^\pi \frac{(\pi-x) \sin x}{\cos ^2 x+3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { On adding Eq. (i) and (ii), we get }\\ &\begin{aligned} 2 I & =\int_0^\pi \frac{\pi \sin x}{3+\cos ^2 x} d x \\ & =\pi \int_0^\pi \frac{\sin x}{3+\cos ^2 x} d x=-\pi \int_1^{-1} \frac{1}{3+t^2} d t \\ & =-\pi \frac{1}{\sqrt{3}}\left(\tan ^{-1} \frac{t}{\sqrt{3}}\right)_1^{-1} \\ \Rightarrow \quad 2 I & =\frac{-\pi}{\sqrt{3}}\left[\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right] \\ & =\frac{-\pi}{\sqrt{3}}\left[-\frac{\pi}{6}-\frac{\pi}{6}\right] \\ & =-\frac{\pi}{\sqrt{3}} \times\left(-\frac{2 \pi}{6}\right)=\frac{\pi^2}{3 \sqrt{3}} \\ \Rightarrow \quad I & =\frac{\pi^2}{6 \sqrt{3}} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int\left(\frac{1+x^2}{1+x^4}\right) d x \\ & =\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}-2+2\right)} d x \\ & =\int \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2} d x \end{aligned} $

Let $x-\frac{1}{x}=t$

$ \begin{aligned} & \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t \\ & =\int \frac{1}{t^2+2} d t=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right) \\ & =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right) \text { or } \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) \end{aligned} $

$ \begin{aligned} A & =\int_0^{\infty} \frac{1+x^2}{1+x^4} d x=\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\right]_0^{\infty} \\ & =\frac{1}{\sqrt{2}}\left[\tan ^{-1} \infty-\tan ^{-1}(-\infty)\right] \\ & =\frac{1}{\sqrt{2}} \times 2 \times \frac{\pi}{2} \Rightarrow \frac{\pi}{\sqrt{2}} \\ B & =\int_0^1 \frac{1+x^2}{1+x^4} d x=\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\right]_0^1 \\ & =\frac{1}{\sqrt{2}}\left[\tan ^{-1}(0)-\tan ^{-1}(-\infty)\right]=\frac{1}{\sqrt{2}} \times \frac{\pi}{2} \\ & \Rightarrow A=2 B \end{aligned} $

We have, $I=\int_0^1 \sqrt{\frac{2+x}{2-x}} d x$

Let $\frac{2+x}{2-x}=u$

$ \left(\frac{1}{2-x}+\frac{x+2}{(2-x)^2}\right) d x=d u $

Now, lower limit $=1$

and upper limit $=3$

$ =\int_1^3 \frac{\sqrt{u}}{(u+1)^2} d u $

Now, let $s=\sqrt{u}$

$ \begin{aligned} d s & =\frac{1}{2 \sqrt{u}} \cdot d u=8 \int_1^{\sqrt{3}} \frac{s^2}{\left(s^2+1\right)^2} d s \\ & =8 \int_1^{\sqrt{3}}\left(\frac{1}{s^2+1}-\frac{1}{\left(s^2+1\right)^2}\right) d s \\ & =8\left|\tan ^{-1}(s)\right|_1^{\sqrt{3}}-8 \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s \\ & =8\left(\tan ^{-1}(\sqrt{3})-\tan ^{-1}(1)\right)-8 \end{aligned} $

$ \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s $

$ \begin{aligned} & =8\left(\frac{\pi}{3}-\frac{\pi}{4}\right)-8 \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s \\ & =\frac{2 \pi}{3}-8 \int_{\pi / 4}^{\pi / 3} \cos ^2 p d p \end{aligned} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { [put } s=\tan p] $

$ \begin{aligned} & =\frac{2 \pi}{3}+2-\sqrt{3}+1-\left.4 p\right|_{\pi / 4} ^{\pi / 3} \\ & =\frac{1}{3}(6-3 \sqrt{3}+\pi)=\frac{\pi}{3}+2-\sqrt{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & M=\int_0^{\infty} \frac{\log t}{1+t^3} \text { and } N=\int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} d t \\ & N=\int_{-\infty}^{\infty} \frac{t \cdot e^{2 t}}{1+e^{3 t}} d t \end{aligned} \end{aligned} $

Let $e^t=x$

$e^t \cdot d t=d x$

Lower limit = 0

and Upper limit $=\infty$

$ \begin{aligned} N & =\int_0^{\infty} \frac{\log x \cdot x^2}{1+x^3} \cdot \frac{d x}{x}=\int_0^{\infty} \frac{x \log x}{1+x^3} d x \\ & =\int_0^{\infty} \frac{x \log x}{1+x^3} d x=-\int_0^{\infty} \frac{\log x}{1+x^3} d x \\ \Rightarrow N & =-\int_0^{\infty} \frac{\log t}{1+t^3} d t \Rightarrow N=-M \end{aligned} $

To solve the integral $ I = \int_{-2}^2 \left(4 - x^2\right)^{\frac{5}{2}} \, dx $, we can use a substitution method.

Substitution:

Substitute $ x = 2 \sin u $:

Then $ dx = 2 \cos u \, du $.

Change the limits:

When $ x = -2 $, $ u = -\frac{\pi}{2} $.

When $ x = 2 $, $ u = \frac{\pi}{2} $.

Substitute into the integral:

$ I = \int_{-\pi/2}^{\pi/2} \left(4 - (2\sin u)^2\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

$ = \int_{-\pi/2}^{\pi/2} \left(4 - 4\sin^2 u\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

Simplify:

Since $ 4 - 4\sin^2 u = 4\cos^2 u $, we have:

$ = \int_{-\pi/2}^{\pi/2} \left(4\cos^2 u\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

$ = 32 \int_{-\pi/2}^{\pi/2} \cos^6 u \, du $

Using Reduction Formula:

Solving this integral using the reduction formula yields:

$ I = 64 \int_{-\pi/2}^{\pi/2} \cos^6 u \, du = 20\pi $

Thus, the value of the integral is $ I = 20\pi $.

$ \begin{aligned} &\text { We have, } P=\lim _{x \rightarrow \infty}\\ &\begin{aligned} & \quad\left[\left(1+\frac{1}{n^3}\right)^{1 / n^3}-\left(1+\frac{8}{n^3}\right)^{4 / n^3} \ldots(2)^{1 / n}\right] \\ & \Rightarrow \quad \log p=\lim _{x \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r^2}{n^2} \log \left(1+\frac{r^3}{n^3}\right) \\ & \Rightarrow \quad \log p=\int_0^1 x^2 \log \left(1+x^3\right) d x \\ & \text { Put } 1+x^3=t \Rightarrow 3 x^2 d x=d t \\ & \text { when } n=0, t=1 \text { and } x=1, t=2 \\ & \therefore \log p=\frac{1}{3} \int_1^2 \log t d t=\frac{1}{3}[t \log t-t]_1^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \log p=\frac{1}{3}[2 \log 2-1] \\ & \text { Hence, } p=e^{(2 \log 2-1 / 3)} \end{aligned} $

To solve the integral $ \int_{-5 \pi}^{5 \pi}(1-\cos 2 x)^{\frac{5}{2}} d x $, we start by simplifying the expression inside the integral:

Notice that $ 1 - \cos 2x = 2 \sin^2 x $. Substituting this in, the integral becomes:

$ I = \int_{-5 \pi}^{5 \pi} (1 - \cos 2x)^{\frac{5}{2}} \, dx = \int_{-5 \pi}^{5 \pi} (2 \sin^2 x)^{\frac{5}{2}} \, dx $

Simplify the expression:

$ = \int_{-5 \pi}^{5 \pi} 2^{\frac{5}{2}} \cdot |\sin^5 x| \, dx $

Since $|\sin^5 x| = \sin^5 x$ for the interval $[0, \pi]$ and is symmetrical over $[-\pi, 0]$, the integral becomes:

$ = 8 \sqrt{2} \int_0^{5\pi} \sin^5 x \, dx $

Exploit the periodicity of the sine function to break the integral over one full period:

$ = 40 \sqrt{2} \int_0^{\pi} \sin^5 x \, dx $

Using the reduction formula for $\int_0^\pi \sin^n x \, dx$, the final result of this reduction gives:

$ = \frac{128 \sqrt{2}}{3} $

$I=\int_0^{\pi / 4} \log (1+\tan x) d x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using $\int_0^a f(x) d x=\int_0^a f(a-x) d x$

$ \begin{aligned} I & =\int_0^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\ & =\int_0^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x \\ & \left(\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right) \\ I & =\int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On adding Eq. (i) and Eq. (ii), we get

$ \begin{aligned} 2 I= & \int_0^{\pi / 4} \log (1+\tan x) d x \\ & \quad+\int_0^{\pi / 4} \log \frac{2}{(1+\tan x)} d x \\ = & \int_0^{\pi / 4} \log (1+\tan x) \cdot \frac{2}{(1+\tan x)} d x \\ = & \int_0^{\pi / 4} \log 2 d x=\log 2 \int_0^{\pi / 4} d x \\ = & \log 2 \cdot(x)_0^{\pi / 4}=\log 2 \cdot\left(\frac{\pi}{4}-0\right) \\ 2 I= & \frac{\pi}{4} \log 2 \Rightarrow I=\frac{\pi}{8} \log 2 \end{aligned} $

$ \begin{aligned} I & =\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x \\ & =2 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =2\left[\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x\right] \\ & =2\left[\pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}-\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\right] \\ 2 I & =2 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned}\\ &\text { }\\ &\begin{aligned}Let\,\,\,\,\, t & =\cos x \\ d t & =-\sin x d x \\ & =-\pi \int_1^{-1} \frac{1}{1+t^2} d t=\pi \int_{-1}^1 \frac{1}{1+t^2} d t \\ I & =\pi\left[\tan ^{-1}(t)\right]_{-1}^1 \\ & =\pi\left[\frac{\pi}{4}+\frac{\pi}{4}\right]=\frac{\pi^2}{2} \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} & \quad I=\int_0^{\pi / 4} \frac{x^2 d x}{(x \sin x+\cos x)} \Rightarrow I=\int_0^{\pi / 4} \frac{x \sec x \cdot \cos x \cdot d x}{(x \sin x+\cos x)^2} \\ & \Rightarrow \quad I=\int_0^{\pi / 4} x \sec x \cdot \frac{x \cos x}{(x \sin x+\cos x)^2} d x \end{aligned} $

Let $I_1=x \sec x$ and $I_2=\frac{x \cos x}{x \sin x+\cos x}$

Let $x \sin x+\cos x=t$

$ \begin{aligned} & x \cos x+\sin x-\sin x=\frac{d t}{d x} \\ & x \cos x d x=d t \end{aligned} $

$ \begin{array}{ll} \Rightarrow & I_2=\int \frac{x \cos x}{(x \sin x+\cos x)^2}=\int \frac{d t}{t^2}=-\frac{1}{t} \\ \Rightarrow \quad & I_2=\frac{-1}{(x \sin x+\cos x)} \\ \Rightarrow \quad & I=\int_0^{\pi / 4} x \sec x \cdot \frac{x \cos x}{(x \sin x+\cos x)} d x \\ \Rightarrow \quad & I=\left[x \sec x \cdot \frac{-1}{(x \sin x+\cos x)}\right]_0^{\frac{\pi}{4}}- \int_0^{\pi / 4}[\sec x+x \sec x \tan x] \cdot \frac{-1}{(x \sin x+\cos x)} d x \\ \because \quad & \int f(x) g(x) d x=f(x) \int g(x) d x-\int\left[\frac{d}{d x} f(x)-\int g(x) d x\right] d x \end{array} $

$ \begin{array}{ll} \Rightarrow & I=\left[\frac{-x \sec x}{x \sin x+\cos x}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \frac{\sec x(1+x \tan x) d x}{(x \sin x+\cos x)} \\ \Rightarrow & I=\left[\frac{-x \sec x}{x \sin x+\cos x}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \frac{\sec ^2 x(\cos x+x \sin x)}{(x \sin x+\cos x)} d x \\ \Rightarrow & I=\left[\frac{-x \sec x}{(x \sin x+\cos x)}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \sec ^2 x d x \end{array} $

$ \begin{array}{ll} \Rightarrow & I=\frac{-\frac{\pi}{4} \cdot \sqrt{2}}{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}+\int_0^{\pi / 4} \sec ^2 x d x \\ \Rightarrow & I=\frac{-\pi \sqrt{2} / 4}{(\pi+4) / 4 \sqrt{2}}+[\tan x]_0^{\pi / 4} \\ \Rightarrow & I=\frac{-2 \pi}{\pi+4}+(1-0)=\frac{-2 \pi}{\pi+4}+1 \Rightarrow I=\frac{4-\pi}{\pi+4} \end{array} $

We have, $I=\int_0^1 \frac{x}{(1-x)^{\frac{3}{4}}} d x$

Let $1-x=t \Rightarrow x=1-t \Rightarrow-d x=d t$

$ \begin{aligned} & \therefore \quad I=\int_0^1 \frac{1-t}{t^{3 / 4}} d t \Rightarrow I=\int_0^1 t^{-3 / 4}(1-t) d t \\ & \Rightarrow \quad I=\int_0^1 t^{-3 / 4} d t-\int_0^1 t^{1 / 4} d t \Rightarrow I=\left[4 t^{1 / 4}\right]_0^1-\left[\frac{4 t^{5 / 4}}{5}\right]_0^1 \end{aligned} $

Putting the value of $t$

$ I=[4(1-x)]_0^1+\left[\frac{4}{5}(1-x)^{5 / 4}\right]_0^1 $

On solving, we get

$ I=\frac{16}{5} $

We have,

$ I=\int_{-1}^1\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) d x $

Let $f(x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$

$ \begin{aligned} & \Rightarrow f(-x)=\sqrt{1-x+x^2}-\sqrt{1+x+x^2} \\ & \Rightarrow f(-x)=-\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) \Rightarrow f(-x)=-f(x) \end{aligned} $

$\because f(x)$ is an odd function.

As, we know that

$ \int_{-a}^a f(x) d x=\left\{\begin{array}{l} \int_0^{2 a} f(x) d x, \text { if } f(-x)=f(x) \text {, even function } \\ 0, \text { if } f(-x)=-f(x), \text { odd funciton } \end{array}\right\} $

$ \begin{aligned} & \text { Then, } I=\int_{-1}^1 \sqrt{1+x+x^2} \\ & \qquad-\sqrt{1-x+x^2} d x=0 \end{aligned} $

We have,

$ \Rightarrow \quad I=\int_1^5|x-3|+|1-x| d x $

Let $f(x)=|x-3|+|1-x|$

Here, when $1 \leq x \leq 3$

$ \Rightarrow f(x)=-x-3-1+x=2 $

When, $3 \leq x \leq 5$

$ \begin{aligned} \Rightarrow f(x) & =x-3-1+x=2 x-4 \\ \Rightarrow \quad I & =\int_1^5|x-3|+|1-x| d x \\ & =\int_1^3 2 d x+\int_3^5(2 x-4) d x \\ \Rightarrow \quad I & =[2 x]_1^3+\left[\frac{2 x^2}{2}-4 x\right]_3^5 \end{aligned} $

On solving, we get

$ I=12 $

If $729 \int_1^3 \frac{1}{x^3\left(x^2+9\right)^2} d x=a+\log b$  Then, $a-b$

On substituting $u=x^2$

$ \begin{array}{ll} \Rightarrow & d u=2 x d x \\ \Rightarrow & d x=\frac{1}{2} \frac{d u}{\sqrt{u}} \\ \text { when, } x=1 & \end{array} $

$ \begin{array}{ll} \Rightarrow & u=1 \\ & x=3 \\ \Rightarrow & u=9 \end{array} $

$ \begin{aligned} &\begin{aligned} I & =729 \int_1^9 \frac{1}{u^{\frac{3}{2}}(u+9)^2} d u \\ & =729 \int_1^9 \frac{1}{2 u^{\frac{3}{2}+\frac{1}{2}}(u+9)^2} d u \\ & =729 \int_1^9 \frac{1}{2 u^2(u+9)^2} d u \end{aligned}\\ &\text { Partial fraction }\\ &\frac{1}{u^2(u+9)^2}=\frac{A}{u}+\frac{B}{u^2}+\frac{C}{(u+9)}+\frac{D}{(u+9)} \end{aligned} $

$ \begin{aligned} &\begin{gathered} \frac{1}{u^2(u+9)^2} \frac{A\left(u(u+9)^2+B(u+9)^2+C u^2(u+9)+D u^2\right.}{u^2(u+9)^2} \\ 1=A u\left(u^2+81+18 u\right)+B\left(u^2+81+18 u\right)+C u^3+9 C u^2+D u^2 \\ 1=A u^3+81 A u+18 A u^2+B u^2+81 B+18 u B+C u^3+9 C u^2+D u^2 \\ 1=(A+C) u^3+(18 A+B+9 C+D) u^2+(81 A+18 B) u+81 B \end{gathered}\\ &\text { Comparison of coefficient } u^3, u^2, u \text { and } 1 \text {. }\\ &\begin{aligned} & A+C=0,18 A+B+9 C+D=0 \\ & 81 A+18 B=0 \Rightarrow 81 B=1 \\ & B=\frac{1}{81} \Rightarrow 81 A+18 \cdot \frac{1}{81}=0 \\ & 81 A=-\frac{2}{9} \Rightarrow A=\frac{-2}{729} \\ & \frac{-2}{729}+C=0 \Rightarrow C=\frac{2}{729} \\ & 18\left(\frac{-2}{729}\right)+\frac{1}{81}+9\left(\frac{2}{729}\right)+D=0 \\ & -\frac{4}{81}+\frac{1}{81}+\frac{2}{81}+D=0 \Rightarrow \frac{-1}{81}+D=0 \end{aligned} \end{aligned} $

$ \begin{aligned} D & =\frac{1}{81} \\ & =\frac{729}{2} \int_1^9\left(\left(\frac{-2}{729}\right) \frac{1}{u}+\frac{1}{81 u^2}+\frac{2}{729(u+9)}+\frac{1}{81(u+9)^2}\right) d u \\ & =\frac{729}{2}\left[\frac{-2}{729} \log u+\frac{1}{81} \frac{u^{-2+1}}{(-2+1)}+\frac{2}{729} \log (u+9)+\frac{1}{81} \frac{(u+9)^{-2+1}}{-2+1}\right]_1^9 \\ & =\frac{729}{2}\left[\frac{-2}{729}(\log 9-\log 1)+\frac{1}{81}(-1)\left(\frac{1}{9}-1\right)+\frac{2}{729}[\log (18)-\log (10)]\right. \\ & =\frac{729}{2}\left[\frac{-2}{729} \log 9-\frac{1}{81}\left(\frac{-8}{9}\right)+\frac{2}{729} \log \left(\frac{18}{10}\right)-\frac{1}{81}\left(\frac{-8}{180}\right)\right]+C \\ & =\frac{1}{2}\left[-2 \log 9+8+2 \log \left(\frac{1}{5}\right)+\frac{8}{20}\right]+C \\ & =\frac{1}{2}\left[-2 \log 9+8+2 \log 9-2 \log 5+\frac{2}{5}\right]+C \\ & =\frac{1}{2}\left[8+\frac{2}{5}-2 \log 5\right]+C \\ & =\frac{1}{2}\left[\frac{40+2}{5}-2 \log 5\right] \Rightarrow \frac{42}{2 \times 5}-\log 5 \end{aligned} $

$ \begin{aligned} a & =\frac{42}{2 \times 5}, b=\frac{1}{5} \\ a & =\frac{21}{5}, b=\frac{1}{5} \\ a-b & =\frac{21}{5}-\frac{1}{5}=\frac{20}{5}=4 \end{aligned} $

To find the limit

$ \lim\limits_{n \rightarrow \infty} \frac{1^{77}+2^{77}+\ldots+n^{77}}{n^{78}} $

we use the method of integral approximation to approximate the sum of powers. Consider the sum

$ S_n = 1^{77} + 2^{77} + \ldots + n^{77}. $

This can be approximated using the integral

$ S_n \approx \int_1^n x^{77} \, dx. $

Evaluating this integral, we have

$ \int_1^n x^{77} \, dx = \left[\frac{x^{78}}{78}\right]_1^n = \frac{n^{78}}{78} - \frac{1}{78}. $

As $ n $ becomes very large, the term $\frac{1}{78}$ becomes negligible, and we approximate

$ S_n \approx \frac{n^{78}}{78}. $

Thus, the given limit can be rewritten and evaluated as follows:

$ \lim_{n \rightarrow \infty} \frac{S_n}{n^{78}} = \lim_{n \rightarrow \infty} \frac{\frac{n^{78}}{78}}{n^{78}} = \lim_{n \rightarrow \infty} \frac{1}{78} = \frac{1}{78}. $

Therefore, the value of the limit is

$ \frac{1}{78}. $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } f(x)=\left\{\begin{array}{cc} \frac{6 x^2+1}{4 x^3+2 x+3}, & 0< x < 1 \\ x^2+1 & 1 \leq x < 2 \end{array}\right. \\ & \int_0^2 f(x) d x=\int_0^1\left(\frac{6 x^2+1}{4 x^3+2 x+3}\right) d x+\int_1^2\left(x^2+1\right) d x \\ & \text { Let } u=4 x^3+2 x+3 \\ & \qquad d u=\left(12 x^2+2\right) d x \Rightarrow 2\left(6 x^2+1\right) d x \\ & \Rightarrow d x=\frac{d u}{2\left(6 x^2+1\right)} \\ & \quad x=0 \Rightarrow u=3 \\ & \quad x=1 \Rightarrow u=4+2+3=9 \\ & \text { Or }\left(6 x^2+1\right) d x=\frac{d u}{2} \Rightarrow \int_3^9 \frac{1}{u}\left(\frac{d u}{2}\right)+\int_1^2\left(x^2+1\right) d x \end{aligned} \end{aligned} $

$ \begin{aligned} & \frac{1}{2}[\log u]_3^9+\left[\frac{x^3}{3}+x\right]_1^2 \\ & \frac{1}{2}[\log 9-\log 3]+\left[\frac{8}{3}+2-\frac{1}{3}-1\right] \\ & \frac{1}{2}\left[\log \left(\frac{9}{3}\right)+\left(\frac{8+6}{3}-\frac{4}{3}\right)\right] \\ & \frac{1}{2} \log 3+\frac{14}{3}-\frac{4}{3}=\frac{1}{2} \log 3+\frac{10}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \int_1^n[x] d x=120 \\ & =\int_1^2 1^{d x}+\int_2^3 2^{d x}+\ldots+\int_{n-1}^n(n-1) d x=120 \\ & \Rightarrow \quad[x]_1^2+[2 x]_2^3+\ldots[(n-1) x]_{n-1}^n=120 \\ & \Rightarrow \quad 2-1+6-4+\ldots+(n-1)=120 \\ & \Rightarrow \quad 1+2+\ldots+n-1=120 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{n(n-1)}{2}=120 \\ & \Rightarrow \quad n^2-n-240=0 \\ & \Rightarrow \quad n^2-16 n+15 n-240=0 \\ & \Rightarrow \quad n(n-16)+15(n-16)=0 \\ & (n-16)(n+15)=0 \\ & n=16 \end{aligned} $

$\int\limits_{-1 / 24}^{1 / 24} \sec x \log \left(\frac{1-x}{1+x}\right) d x$

Here, $f(x)=\sec x \log \left(\frac{1-x}{1+x}\right)$

Now, $f(-x)=\sec (-x) \log \left(\frac{1-(-x)}{1-x}\right)$

$ \begin{aligned} & =\sec x \log \left(\frac{1+x}{1-x}\right) \\ & =-\sec x \log \left(\frac{1-x}{1+x}\right)=-f(x) \end{aligned} $

Thus, the given function is an odd function

$\therefore \quad \int\limits_{\frac{-1}{24}}^{\frac{1}{24}} \sec x \log \left(\frac{1-x}{1+x}\right) d x=0$

$\left[\because \int\limits_{-a}^a f(x) d x=0\right.$, if $f(x)$ is an odd function]

$I=\int_0^5[x] d x$

$ =\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x+\int_3^4 3 d x+\int_4^5 4 d x $

$=0+(2-1)+2(3-2)+3(4-3)+4(5-4)$ $=1+2+3+4=10$

$I=\int_0^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x\quad \ldots \text { (i) } $

$ \begin{aligned} & =\int_0^{\pi / 2} \frac{1}{\sqrt{1+\tan \left(\frac{\pi}{2}-x\right)}} d x \\ & =\int_0^{\pi / 2} \frac{1}{\sqrt{1+\cot x}} d x \\ & =\int_0^{\pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x \quad \ldots \text { (ii) } \end{aligned} $

$\therefore$ From Eqs. (i) and (ii), we get

$ 2 I=\int_0^{\pi / 2} 1 d x=[x]_0^{\pi / 2}=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} $

To solve the integral $ I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} \, dx $, we can use symmetry properties of definite integrals. According to the property:

$ \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx $

Apply this property to transform the integral:

$ I = \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} \, dx $

Adding both expressions for $ I $, we have:

$ 2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} \, dx $

Now, perform substitution to evaluate the new integral. Let $\cos x = t$, then $ -\sin x \, dx = dt $. The limits of integration change as $ x $ changes from 0 to $\pi$, causing $ t $ to vary from 1 to -1. Thus, we have:

$ -\pi \int_1^{-1} \frac{dt}{1+t^2} $

Reversing the limits, the expression becomes:

$ \pi \int_{-1}^{1} \frac{dt}{1+t^2} $

This integral evaluates to:

$ \pi \left[ \tan^{-1} t \right]_{-1}^{1} $

Simplifying:

$ = \pi \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) $

Since $\tan^{-1}(1) = \frac{\pi}{4}$, it follows that:

$ = \pi \left( \frac{\pi}{4} + \frac{\pi}{4} \right) = \pi \times \frac{\pi}{2} = \frac{\pi^2}{2} $

Thus, the original integral $ I $ is:

$ I = \frac{1}{2} \times \frac{\pi^2}{2} = \frac{\pi^2}{4} $

We have $I=\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$

Let $f(x)=\frac{x \sin ^3 x}{4-\cos ^2 x}$

$ f(-x)=\frac{-x \times \sin ^3(-x)}{4-\cos ^2(-x)}=\frac{x \sin ^3 x}{4-\cos ^2 x}=f(x) $

$\therefore f(x)$ is an even function

$ \begin{aligned} & I=2 \int_0^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x \\ & I=2 \int_0^\pi \frac{(\pi-x) \sin ^3 x}{4-\cos ^2 x} d x \end{aligned} $

On adding Eqs. (ii) and (ii), we get

$ \begin{aligned} & 2 I=2 \pi \int_0^\pi \frac{\sin ^3 x}{4-\cos ^2 x} d x \\ & I=\pi \int_0^\pi \frac{\sin ^3 x}{4-\cos ^2 x} d x \end{aligned} $

Let, $\cos x=t \Rightarrow-\sin x d x=d t$

When $x \rightarrow 0 t \rightarrow 1$ and $x \rightarrow \pi, t \rightarrow-1$

$ \begin{aligned} I & =-\pi \int_1^{-1} \frac{\left(1-t^2\right)}{4-t^2} d t \\ & =\pi \int_{-1}^1\left(\frac{-3}{4-t^2}+1\right) d t \\ & =\pi\left[\frac{-3}{4} \log \left|\frac{2+t}{2-t}\right|+t\right]_{-1}^1 \\ & =\pi\left[\left(\frac{-3}{4} \log 3+1\right)-\left(-\frac{3}{4} \log \frac{1}{3}-1\right)\right] \\ & =\pi\left[-\frac{3}{4} \log 3+1-\frac{3}{4} \log 3+1\right] \\ & =2 \pi\left[1-\frac{3}{4} \log 3\right] \end{aligned} $

To solve the integral, we have:

$ I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{\sqrt[5]{x^{30} + x^{25}}} \, dx $

First, we can rewrite the integrand as:

$ \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{x^6 \sqrt[5]{1 + \frac{1}{x^5}}} \, dx $

To simplify, we perform a substitution. Let:

$ 1 + \frac{1}{x^5} = t^5 $

Differentiating both sides gives:

$ -\frac{5}{x^6} \, dx = 5t^4 \, dt $

Thus, the differential becomes:

$ dx = -x^6 \cdot \frac{1}{5t^4} \, dt $

We need to adjust the limits of integration. When $ x \rightarrow \frac{1}{\sqrt[5]{31}} $, then $ t \rightarrow 2 $. When $ x \rightarrow \frac{1}{\sqrt[5]{242}} $, then $ t \rightarrow 3 $.

The integral now becomes:

$ I = -\int_2^3 \frac{1}{t} \times t^4 \, dt $

This simplifies to:

$ I = -\int_2^3 t^3 \, dt $

$ I = -\left[ \frac{t^4}{4} \right]_2^3 $

Calculating the definite integral gives:

$ I = -\left( \frac{3^4}{4} - \frac{2^4}{4} \right) $

$ I = -\left( \frac{81}{4} - \frac{16}{4} \right) $

$ I = -\left( \frac{65}{4} \right) $

Thus, the evaluated integral is:

$ I = \frac{-65}{4} $

$ \begin{aligned} &\text {Let, }\\ &\begin{aligned} & P=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right)\right. \\ & \left.\ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n} \\ & \log P=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\log \left(1+\frac{1}{n^2}\right)\right. \\ & \left.+\log \left(1+\frac{2^2}{n^2}\right)+\ldots \log \left(1+\frac{n^2}{n^2}\right)\right] \\ & \log P=\lim _{n \rightarrow-\infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}\right) \\ & =\int_0^1 \log \left(1+x^2\right) d x \\ & =\left[x \log \left(1+x^2\right)\right]_0^1-\int_0^1 \frac{2 x^2}{1+x^2} d x \\ & =\log 2-2 \int_0^1 d x+2 \int_0^1 \frac{x^2+1-1}{1+x^2} d x \\ & =\log 2-2 \int_0^1 d x+2 \int_0^1 \frac{d x}{1+x^2} \\ & =\log 2-z x]_0^1+2\left[\tan ^{-1} a\right]^{0^1} \\ & \log P=\log 2-2+\frac{\pi}{2} \\ & P=e^{\log 2-2+\frac{\pi}{2}}=e^{\log 2} \cdot e^{-2+\frac{\pi}{2}} \\ & P=2 e^{-2+\frac{\pi}{2}} \\ & \therefore \quad a=2 \quad b=-2+\frac{\pi}{2} \\ & a+b=2-2+\frac{\pi}{2}=\frac{\pi}{2} \end{aligned} \end{aligned} $

Let $I=\int_0^\pi x \sin ^4 x \cos ^6 x d x...(i)$

$ I=\int_0^\pi(\pi-x) \sin ^4 x \cos ^6 x d x ...(ii)$

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & I=\pi \int_0^\pi \sin ^4 x \times \cos ^6 x d x \\ & I=\frac{\pi}{2} \times 2 \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x \\ & I=\pi \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x \end{aligned} $

$=\frac{\pi\left[\frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi} \times \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi}\right]}{2 \times 120}=\frac{3 \pi^2}{512}$

To determine $ I_n $, we start with the given integral:

$ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx $

Rewrite this as:

$ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \times (\sec^2 x - 1) \, dx $

Breaking it down, we have:

$ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx $

Therefore, we can express this as:

$ I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx $

When integrating, we find:

$ \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx = \left[ \frac{\tan^{n-1} x}{n-1} \right]_0^{\frac{\pi}{4}} = \frac{1}{n-1} $

Hence, we deduce:

$ I_n + I_{n-2} = \frac{1}{n-1} $

By setting $ n = 13 $, it follows that:

$ I_{13} + I_{11} = \frac{1}{12} $

Thus, the sum $ I_{13} + I_{11} $ is:

$ \frac{1}{12} $

$ \begin{aligned} & \lim _{n \rightarrow \infty} x^4\left[\begin{array}{l} \frac{1}{x^5}+\frac{1}{\left(x^2+1\right)^{5 / 2}}+\frac{1}{\left(x^2+4\right)^{5 / 2}} \\ +\frac{1}{\left(x^2+9\right)^{5 / 2}}+\ldots+\frac{1}{4 \sqrt{2} x^5} \end{array}\right] \\ & \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{x^4}{\left(x^2+r^2\right)^{5 / 2}}\left[\because \lim _{n \rightarrow \infty} \frac{x^4}{x^5}=0\right] \\ & \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left[\frac{1}{\left[1+\left(\frac{r}{n}\right)^2\right]^{5 / 2}}\right] \\ & =\int_0^1 \frac{1}{\left(1+x^2\right)^{5 / 2}} d x \\ & \text { Let } x=\tan \theta \\ & d x=\sec ^2 \theta d \theta \end{aligned} $

$\begin{aligned} x \rightarrow 0, \theta & \rightarrow 0 \text { and } \\ x \rightarrow 1, \theta & \rightarrow \frac{\pi}{4} \\ & =\int_0^{\pi / 4} \frac{1}{\left(1+\tan ^2 \theta\right)^{5 / 2}} \times \sec ^2 \theta d \theta \\ & =\int_0^{\pi / 4} \frac{1}{\sec ^3 \theta} d \theta \\ & =\int_0^{\pi / 4} \cos ^3 \theta d \theta \\ & =\frac{1}{4} \int_0^{\pi / 4}(\cos 3 \theta+3 \cos \theta) d \theta \\ & =\frac{1}{4}\left[\frac{\sin 3 \theta}{3}+3 \sin \theta\right]_0^{\pi / 4} \\ & =\frac{1}{4}\left[\frac{1}{3} \sin \left(\pi-\frac{\pi}{4}\right)+3 \sin \frac{\pi}{4}\right] \\ & =\frac{1}{4}\left[\frac{1}{3} \times \frac{1}{\sqrt{2}}+\frac{3}{\sqrt{2}}\right] \\ & =\frac{1+9}{4 \times 3 \sqrt{2}} \Rightarrow \frac{10}{4 \times 3 \sqrt{2}}=\frac{5}{6 \sqrt{2}}\end{aligned}$

$\begin{aligned} \text {Let} \quad I & =\int_{\log 4}^{\log 5} \frac{e^{2 x}+e^x}{e^{2 x}-5 e^x+6} d x \\ & =\int_{\log 4}^{\log 5} \frac{\left(e^x\right)^2+e^x}{\left(e^x-3\right)\left(e^x-2\right)} d x \\ & =\int_{\log 4}^{\log 5} \frac{e^x\left(e^x+1\right)}{\left(e^x-3\right)\left(e^x-2\right)} d x\end{aligned}$

Let $e^x=t$

$ \begin{aligned} & \Rightarrow \quad e^x d x=d t \\ & \text { When } x \rightarrow \log 4, t \rightarrow 4 \end{aligned} $

$ \begin{aligned} & x \rightarrow \log 5, t \rightarrow 5 \\ & I=\int_4^5 \frac{t+1}{(t-3)(t-2)} d t \\ & \quad=\int_4^5 \frac{4}{t-3}-\frac{3}{t-2} \end{aligned} $

[ $\because$ by partial fraction]

$\begin{aligned} & =[4 \log |t-3|-3 \log |t-2|]_4^5 \\ & =\left[\log \left\lvert\, \frac{(t-3)^4}{(t-2)^3}\right.\right]_4^5 \\ & =\log \frac{2^4}{3^3}-\log \frac{1}{2^3}=\log \frac{2^4}{3^3} \times 2^3 \\ & =\log \left(\frac{2^7}{3^3}\right)=\log \left(\frac{128}{27}\right)\end{aligned}$

$\begin{aligned} & \text { Let } I=\int_1^2 \frac{x^4-1}{x^6-1} d x \\ & \quad=\int_1^2 \frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^2-1\right)\left(x^4+x^2+1\right)} d x \\ & =\int_1^2 \frac{x^2+1}{x^4+x^2+1} d x=\int_1^2\left(\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}\right) d x \\ & =\int_1^2\left(\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-2+3}\right) d x \\ & =\int_1^2 \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+3} d x\end{aligned}$

Let $x-\frac{1}{x}=t \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t$

When $x \rightarrow 1$, then $t \rightarrow 0$

When $x \rightarrow 2$, then $t \rightarrow \frac{3}{2}$

$ \begin{aligned} I & =\int_0^{3 / 2} \frac{1}{t^2+3} d t \\ & =\frac{1}{\sqrt{3}}\left[\tan ^{-1} \frac{1}{\sqrt{3}}\right]_0^{3 / 2} \\ & =\frac{1}{\sqrt{3}} \tan ^{-1} \frac{3}{2 \sqrt{3}} \\ & =\frac{1}{\sqrt{3}} \tan ^{-1} \frac{\sqrt{3}}{2} \end{aligned} $

Let $ I=\int_{0}^{\infty} \frac{1}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x $

$ \begin{array}{l} I=\frac{1}{a^{2}-b^{2}} \int_{0}^{\infty} \frac{\left(x^{2}+a^{2}\right)-\left(x^{2}+b^{2}\right)}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x \\ I=\frac{1}{a^{2}-b^{2}} \int_{0}^{\infty} \frac{1}{\left(x^{2}+b^{2}\right)}-\frac{1}{\left(x^{2}+a^{2}\right)} d x \\ I=\frac{1}{a^{2}-b^{2}}\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}-\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]_{0}^{\infty} \\ I=\frac{1}{a^{2}-b^{2}}\left[\frac{1}{b} \times \frac{\pi}{2}-\frac{1}{a} \times \frac{\pi}{2}\right] \\ I=\frac{1}{(a+b)(a-b)}\left[\frac{a-b}{a b}\right] \times \frac{\pi}{2} \\ I=\frac{\pi}{2 a b(a+b)} \end{array} $

Let $ I=\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x \quad \ldots \text { (i) } $

By applying property,

$ \begin{aligned} \int_{a}^{b} f(x) d x & =\int_{a}^{b} f(a+b-x) d x \\ I & =\int_{0}^{\frac{\pi}{2}} \log \left(\tan \left(\frac{\pi}{2}-x\right)\right) d x \\ I & =\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x \end{aligned} \quad \ldots \text { (ii) } $

Adding Eqs. (i) and (ii),

$ \begin{array}{l} 2 I=\int_{0}^{\frac{\pi}{2}}[\log (\tan x)+\log (\cot x)] d x \\ 2 I=\int_{0}^{\frac{\pi}{2}} \log 1 d x=0 \end{array} $

1. The given integral is:

$ I=\int_0^1 \frac{dx}{\left(5+2x-2x^2\right)\left(1+e^{2-4x}\right)} $ .............(i)

1. Perform a substitution, $x \rightarrow 1-x$. This gives:

$ I=\int_0^1 \frac{dx}{\left(5+2(1-x)-2(1-x)^2\right)\left(1+e^{2-4(1-x)}\right)} $

1. Simplify the expression inside the integral:

$ I=\int_0^1 \frac{dx}{\left(5+2-2x-2(1-2x+x^2)\right)\left(1+e^{4x -2}\right)} $ ............(ii)

1. Add the original integral (i) and the integral after substitution (ii):

$ 2I=\int_0^1 \frac{dx}{5+2x-2x^2} $

1. Factor the quadratic expression in the denominator:

$ 2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)} $

1. To solve this integral, we can perform a change of variables using the substitution
   
   $x-\frac{1}{2}= \frac{\sqrt{11}}{2}\tan{u}$, then $dx=\frac{\sqrt{11}}{2}\sec^2{u} du$:

$ I=\frac{1}{\sqrt{11}}\int \frac{\sec^2{u}}{1+\tan^2{u}}du $

1. Using the identity $\sec^2{u}=1+\tan^2{u}$:

$ I=\frac{1}{\sqrt{11}}\int du = \frac{1}{\sqrt{11}}(u+C) $

1. Now we need to find the limits of the integral after the substitution. If $x=0$, then $u=\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)$. If $x=1$, then $u=\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)$. So, the integral becomes:

$ I=\frac{1}{\sqrt{11}}\left[\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right] $

1. Using the properties of the arctangent function, we can rewrite the integral as:

$ I=\frac{1}{\sqrt{11}} \ln\left(\frac{\sqrt{11}+1}{\sqrt{10}}\right) $

1. From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$. Now, we can find $\alpha^4 - \beta^4$:

$ \alpha^4 - \beta^4 = (11)^2 - (10)^2 = 121 - 100 = 21 $

Thus, $\alpha^4 - \beta^4 = 21$.

We're given the expression:

$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$

Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan^n x$ from 0 to $\pi / 4$, where $n$ is an integer. Let's denote this integral as $I(n)$:

$I(n) = \int_0^{\pi / 4} e^{-x} \tan^n x dx$

We can then rewrite the original expression in terms of $I(n)$:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$

Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:

$\int u dv = uv - \int v du$

We'll choose:

$u = \tan^n x, \quad dv = e^{-x} dx$

Then we get:

$du = n \tan^{n-1} x \sec^2 x dx, \quad v = -e^{-x}$

Applying integration by parts, we have:

$I(n) = -e^{-x} \tan^n x \Bigg|_0^{\pi / 4} + n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x \sec^2 x dx$

Since $\tan(\pi / 4) = 1$, the first term evaluates to:

$-e^{-\pi / 4}$

The second term becomes:

$n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) dx = n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x dx + n \int_0^{\pi / 4} e^{-x} \tan^{n+1} x dx$

This is equal to:

$n(I(n-1) + I(n+1))$

So we have:

$I(n) = -e^{-\pi / 4} + n(I(n-1) + I(n+1))$

Now we can substitute $n = 50$ into this equation:

$I(50) = -e^{-\pi / 4} + 50(I(49) + I(51))$

So the original expression becomes:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)} = \frac{e^{-\pi/4} - e^{-\pi / 4} + 50(I(49) + I(51))}{I(49) + I(51)} = 50$