Definite Integration

$ \begin{aligned} & \alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x ........(i) \\ & \text { Let } x=\frac{1}{t} \\ & \qquad d x=-\frac{1}{t^2} d t \\ & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t \end{aligned} $

$ \begin{aligned} & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t \\ & \alpha=\int_{\frac{1}{2}}^2 \frac{\cot ^{-1} t}{2 t^2-3 t+2} d t .......(ii) \end{aligned} $

Now by (i) + (ii)

$ \begin{aligned} & 2 \alpha=\int_{\frac{1}{2}}^2 \frac{\frac{\pi}{2}}{2 x^2-3 x+2} d x \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{x^2-\frac{3 x}{2}+1} \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{\left(x-\frac{3}{4}\right)^2+\frac{7}{16}} \\ & \alpha=\frac{\pi}{8 \times \frac{\sqrt{7}}{4}}\left[\tan ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right)\right]_{\frac{1}{2}}^2 \end{aligned} $

$\begin{aligned} & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{4 x-3}{\sqrt{7}}\right]_{\frac{1}{2}}^2 \\ & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right] \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1} \frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}} \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1}(3 \sqrt{7}) \end{aligned}$

Now $\sqrt{7} \tan \left(\frac{2 \sqrt{7} \alpha}{\pi}\right)$

$ \begin{aligned} & \sqrt{7} \times \tan \left(\tan ^{-1}(3 \sqrt{7})\right) \\ & \sqrt{7} \times 3 \sqrt{7} \\ & =21 \end{aligned} $

$\mathrm{I}=2 \int_\limits0^{\frac{\pi}{2}} \underbrace{\sin ^2 \mathrm{x} \cdot \sqrt{\frac{\pi x}{2}-\mathrm{x}^2}}_{\mathrm{i}_1}-\int_\limits0^{\frac{\pi}{2}} \mathrm{~g}(\mathrm{x}) \mathrm{dx}$

$\text { Let } I_1=\int_\limits0^{\frac{\pi}{2}} \sin ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} \quad \text { (making perfect square) }$

apply kings

$I_1=\int_\limits0^{\frac{\pi}{2}} \cos ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{2}-x\right)^2}$

add both

$2 I_1=\int_\limits0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$

i.e. $2 I_1=\int_\limits0^{\frac{\pi}{2}} g(x)$

Now $I=2 I_1-\int_\limits0^{\frac{\pi}{2}} g(x)=0$

Now $I_1=\int_\limits0^{\frac{\pi}{2}} f(x) \cdot g(x) d x=\frac{1}{2} \int_\limits0^{\frac{\pi}{2}} g(x) d x$

i.e. $\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} d x$

Using $\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C$

$\begin{aligned} & \Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{\pi^2}{16}}{2} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2} \\ & \Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right] \\ & \Rightarrow \frac{1}{2} \times \frac{\pi^3}{32} \end{aligned}$

Now $\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25$

$\begin{aligned} f(x) & =\int\limits_0^{x \tan ^{-1} x} \frac{e^{t-\cos t}}{1+t^{2023}} d t \\\\ f^{\prime}(x) & =\frac{e^{x \tan ^{-1} x-\cos \left(x \tan ^{-1} x\right)}}{1+\left(x \tan ^{-1} x\right)^{2023}}\left(\frac{x}{1-x^2}+\tan ^{-1} x\right)\end{aligned}$

For max/min put $f^{\prime}(x)=0$

$ \begin{aligned} & \Rightarrow \frac{x}{1+x^2}+\tan ^{-1} x=0 \\\\ & \Rightarrow x=0 \end{aligned} $


$\therefore f(x)$ has minimum at $x=0$

And $f(x)_{\min }=f(0)=0$

Let $I = \int_1^2 {{{\log }_2}({x^3} + 1)dx + \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx} } $

Let ${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx} $

and ${I_2} = \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx} $

Let ${2^x} - 1 = {y^3}$

$ \Rightarrow {2^x}\,.\,\log _2^2 = 3{y^2}\,.\,{{dy} \over {dx}}$

$ \Rightarrow dx = {{3{y^2}} \over {{2^x}}}dy$

$ \Rightarrow dx = {{3{y^2}dy} \over {{y^3} + 1}}$

When $x = 1$ then ${y^3} = {2^1} - 1 = 1 \Rightarrow y = 1$

When $x = \log _2^9$ then ${y^3} = {2^{\log _2^9}} - 1 \Rightarrow {y^3} = 8 \Rightarrow y = 2$

$\therefore$ ${I_2} = \int_1^2 {y\,.\,{{3{y^2}} \over {{y^3} + 1}}dy} $

$ = \int_1^2 {{{3{y^3}dy} \over {{y^3} + 1}}} $

Let $y = x$

$ \Rightarrow dy = dx$

$\therefore$ ${I_2} = \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}} $

Now, ${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx} $

$ = \left[ {{{\log }_2}({x^3} + 1)\,.\,x} \right]_1^2 - \int_1^2 {{1 \over {{x^3} + 1}}\,.\,{{3{x^2}} \over {\log _2^2}}\,.\,x\,dx} $

$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_2^2 - \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}} $

$\therefore$ $I = {I_1} + {I_2}$

$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2 - \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}} + \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}}} } $

$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2$

$ = 2\log _2^9 - 1\,.\,\log _2^2$

$ = 2\log _2^9 - 1$

$ = 4\log _2^3 - 1$

$ = 4 \times 1.58 - 1$

$ = 6.32 - 1$

$ = 5.32$

$\therefore$ Greatest integer value of

$I = [5.32] = 5$

Other Method :-

Let $f(x) = {\log _2}({x^3} + 1) = y$

$ \Rightarrow {x^3} + 1 = {2^y}$

$ \Rightarrow {x^3} = {2^y} - 1$

$ \Rightarrow x = {\left( {{2^y} - 1} \right)^{{1 \over 3}}}$

$\therefore$ ${f^{ - 1}}(x) = {\left( {{2^x} - 1} \right)^{{1 \over 3}}}$

And $f(1) = \log _2^{(1 + 1)} = 1 = f(a)$ (Assume)

$f(2) = \log _2^{(8 + 1)} = \log _2^9 = f(b)$

$\therefore$ $I = \int_a^b {f(x)dx + \int_{f(a)}^{f(b)} {{f^{ - 1}}(x)dx} } $

Let ${f^{ - 1}}(x) = t$

$ \Rightarrow x = f(t)$

$ \Rightarrow dx = f'(t)dt$

When $x = f(a)$ then $f(a) = f(t) \Rightarrow t = a$

When $x = f(b)$ then $f(b) = f(t) \Rightarrow t = b$

$\therefore$ $I = \int_a^b {f(t)dt + \int_a^b {t\,.\,f'(t)dt} } $

$ = \int_a^b {\left( {f(t) + t\,.\,f'(t)} \right)dt} $

$ = \left[ {t\,.\,f(t)} \right]_a^b$

$ = b\,.\,f(b) - a\,.\,f(a)$

Here $b = 2$, $f(b) = \log _2^9$

and $a = 1$, $f(a) = 1$

$\therefore$ $I = 2\log _2^9 - 1\,.\,1$

$ = 4\,.\,\log _2^3 - 1$

$ = 4 \times 1.58 - 1$

$ = 6.32 - 1$

$ = 5.32$

$\therefore$ $[I] = 5$

${S_1} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}x.1\,dx = \int_{\pi /8}^{3\pi /8} {\left( {{{1 - \cos 2x} \over 2}} \right)dx} } $

$ = \left( {{x \over 2} - {{\sin 2x} \over 4}} \right)_{\pi /8}^{3\pi /8} = {\pi \over 8}$

$\therefore$ ${{16{S_1}} \over \pi } = {{16} \over \pi } \times {\pi \over 8} = 2$

${S_2} = \int\limits_{\pi /8}^{3\pi /8} {{{\sin }^2}x\left| {4x - \pi } \right|dx} $ .... (i)

${S_2} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right)\left| {4\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right) - \pi } \right|dx} $


$[\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx]} } $

$ \Rightarrow {S_2} = \int_{\pi /8}^{3\pi /8} {{{\cos }^2}x\left| {\pi - 4x} \right|dx} $ .... (ii)

Adding Eqs. (i) and (ii), we get

$2{S_2} = \int_{\pi /8}^{3\pi /8} {\left| {4x - \pi } \right|dx} $

From figure,

${A_1} = {1 \over 2} \times {\pi \over 8} \times {\pi \over 2} = {{{\pi ^2}} \over {32}} = {A_2}$

$\therefore$ $2{S_2} = 2{A_1} = {{{\pi ^2}} \over {16}} \Rightarrow {S_2} = {{{\pi ^2}} \over {32}}$

Hence, ${{48{S_2}} \over {{\pi ^2}}} = {{48} \over {32}} = {3 \over 2} = 1.5$

Given, $I = \int_0^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]d} x$ .... (i)

${{10x} \over {x + 1}} = 1 \Rightarrow 10x = x + 1 \Rightarrow x = {1 \over 9}$

${{10x} \over {x + 1}} = 4 \Rightarrow 10x = 4x + 4 \Rightarrow x = {2 \over 3}$

${{10x} \over {x + 1}} = 9 \Rightarrow 10x = 9x + 9 \Rightarrow x = 9$

From Eq. (i), we get

$I = \int_0^{1/9} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{1/9}^{2/3} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{2/3}^9 {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_9^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx} } } } $

$ = \int_0^{1/9} {0.dx + \int_{1/9}^{2/3} {1.dx + \int_{2/3}^9 {2.dx + \int_9^{10} {3.dx} } } } $

$ = 0 + [x]_{1/9}^{2/3} + [2x]_{2/3}^9 + [3x]_9^{10}$

$ = {2 \over 3} - {1 \over 9} + 18 - {4 \over 3} + 30 - 27 = 21 - {2 \over 3} - {1 \over 9} = {{182} \over 9}$

Hence, 9I = 182

It is given that, for functions

$f:R \to R$

and $F:[0,\pi ] \to R$,

$F(x) = \int_0^x {f(t)dt} $, where $f(\pi ) = - 6$

$ \Rightarrow F'(\pi ) = f(\pi ) = - 6$ .... (i)

Now, $\int_0^\pi {(f'(x)} + F(x))\cos x\,dx$

$ = \int_0^\pi {f'(x)\cos x\,dx + \int_0^\pi {F(x)\cos x\,dx} } $

$ = [\cos x\,f(x)]_0^\pi + \int_0^\pi {f(x)\sin x\,dx} + \int_0^\pi {F(x)\cos x\,dx} $

{by integration by parts}

$ = ( - 1)( - 6) - f(0) + [(\sin x)F(x)]_0^\pi - \int_0^\pi {F(x)\cos x\,dx + \int_0^\pi {F(x)\cos x\,dx} } $

$ = 6 - f(0) = 2$ (given)

$ \Rightarrow f(0) = 4$

The given integral

$I = \int\limits_0^{\pi /2} {{{3\sqrt {\cos \theta } } \over {{{(\sqrt {\cos \theta } + \sqrt {\sin \theta } )}^5}}}} d\theta $ ....(i)

$ \Rightarrow I = \int\limits_0^{\pi /2} {{{3\sqrt {\sin \theta } } \over {{{(\sqrt {\sin \theta } + \sqrt {\cos \theta } )}^5}}}} d\theta $ ....(ii)

[Using the property $\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } $]

Now, on adding integrals (i) and (ii), we get

$2I = \int\limits_0^{\pi /2} {{3 \over {{{(\sqrt {\sin \theta } + \sqrt {\cos \theta } )}^4}}}} d\theta $

$ = \int\limits_0^{\pi /2} {{{3{{\sec }^2}\theta } \over {{{(1 + \sqrt {\tan \theta } )}^4}}}} d\theta $

Now, let $\tan \theta = {t^2} \Rightarrow {\sec ^2}\theta \,d\theta = 2t\,dt$

and at $\theta = {\pi \over 2}$, t $ \to $ $\infty $

and at $\theta = 0$, t $ \to $ 0

So, $2I = \int_0^\infty {{{6\,t\,dt} \over {{{(1 + t)}^4}}} = 6} \int_0^\infty {{{t + 1 - 1} \over {{{(t + 1)}^4}}}} dt$

$ \Rightarrow I = 3\left[ {\int_0^\infty {{{dt} \over {{{(t + 1)}^3}}} - } \int_0^\infty {{{dt} \over {{{(t + 1)}^4}}}} } \right]$

$ = 3\left[ { - {1 \over {2{{(t + 1)}^2}}} + {1 \over {3{{(t + 1)}^3}}}} \right]_0^\infty $

$ \Rightarrow I = 3\left[ {{1 \over 2} - {1 \over 3}} \right] = 3\left( {{1 \over 6}} \right) = {1 \over 2} \Rightarrow I = 0.5$

Given,

$I = {2 \over \pi }\int_{ - \pi /4}^{\pi /4} {{{dx} \over {(1 + {e^{\sin x}})(2 - \cos 2x)}}} $ .... (i)

On applying property

$\int_a^b {f(x)dx} = \int_a^b {f(a + b - x)dx} $, we get

$I = {2 \over \pi }\int_{ - \pi /4}^{\pi /4} {{{{e^{\sin x}}dx} \over {(1 + {e^{\sin x}})(2 - \cos 2x)}}} $ ..... (ii)

On adding integrals (i) and (ii), we get

$2I = {2 \over \pi }\int_{ - \pi /4}^{\pi /4} {{{dx} \over {2 - \cos 2x}}} $

$ \Rightarrow $ $I = {1 \over \pi }\int_{ - \pi /4}^{\pi /4} {{{dx} \over {2 - {{1 - {{\tan }^2}x} \over {1 + {{\tan }^2}x}}}}} $

$\left[ {as\,\cos \,2x = {{1 - {{\tan }^2}x} \over {1 + {{\tan }^2}x}}} \right]$

= ${2 \over \pi }\int_0^{\pi /4} {{{{{\sec }^2}x} \over {1 + 3{{\tan }^2}x}}dx} $

[$ \because $ ${{{{{\sec }^2}x} \over {1 + 3{{\tan }^2}x}}}$ is even function]

Put $\sqrt 3 \tan \,x = t \Rightarrow \sqrt 3 {\sec ^2}dx = dt$, and at x = 0, t = 0 and at x = $\sqrt 3 $, t = $\sqrt 3 $

So, $I = {2 \over \pi }\int_0^{\sqrt 3 } {{1 \over {\sqrt 3 }}{{dt} \over {1 + {t^2}}}} = {2 \over {\sqrt 3 \pi }}[{\tan ^{ - 1}}t]_0^{\sqrt 3 }$

$ = {2 \over {\sqrt 3 \pi }}\left( {{\pi \over 3}} \right) = {2 \over {3\sqrt 3 }} \Rightarrow 27{I^2} = 4.00$

Let $I = \int_0^{1/2} {{{1 + \sqrt 3 } \over {{{[{{(x + 1)}^2}{{(1 - x)}^6}]}^{1/4}}}}dx} $

$ \Rightarrow I = \int_0^{1/2} {{{1 + \sqrt 3 } \over {{{(1 - x)}^2}{{\left[ {{{\left( {{{1 - x} \over {1 + x}}} \right)}^6}} \right]}^{1/4}}}}dx} $

Put ${{1 - x} \over {1 + x}} = t \Rightarrow {{ - 2dx} \over {{{(1 + x)}^2}}} = dt$

when x = 0, t = 1, x = ${1 \over 2}$, t = ${1 \over 3}$

$ \therefore $ $I = \int_1^{1/3} {{{(1 + \sqrt 3 )dt} \over { - 2{{(t)}^{6/4}}}}} $

$ \Rightarrow I = {{ - (1 + \sqrt 3 )} \over 2}\left[ {{{ - 2} \over {\sqrt t }}} \right]_1^{1/3}$

$ \Rightarrow I = (1 + \sqrt 3 )(\sqrt 3 - 1) \Rightarrow I = 3 - 1 = 2$

Let $f(x)=\int_0^x \frac{t^2}{1+t^4} d t-(2 x-1)$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \therefore f^{\prime}(x)=\frac{x^2}{1+x^4}-2 \\\\ & \therefore f^{\prime}(x)<0(\because x \in[0,1]) \end{aligned} $ $\Rightarrow f$ is strictly decreasing function

$ \begin{aligned} \text { Let } \mathrm{I} & =\frac{1}{2} \int_0^x \frac{2 t^2}{t^4+1} d t=\frac{1}{2} \int_0^x \frac{\left(t^2+1\right)+\left(t^2-1\right)}{t^4+1} \\\\ & =\frac{1}{2} \int_0^x \frac{t^2+1}{t^4+1} d t+\frac{1}{2} \int_0^x \frac{t^2-1}{t^4+1} d t \end{aligned} $

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

$ \begin{aligned} & \because f(x)=I-(2 x-1) \\\\ & \therefore f(0)=\frac{1}{2 \sqrt{2}} \tan ^{-1}(-\infty)+\frac{1}{4 \sqrt{2}} \ln (1)+\frac{1}{2 \sqrt{2}}\left(\frac{\pi}{2}\right)+1 \\\\ & \Rightarrow f(0)=\frac{-\pi}{4 \sqrt{2}}+0+\frac{\pi}{4 \sqrt{2}}+1 \\\\ & \Rightarrow f(0)=1 \\\\ & f(1)=\frac{1}{2 \sqrt{2}} \tan ^{-1}(-0)+\frac{1}{4 \sqrt{2}} \ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)+\frac{\pi}{4 \sqrt{2}}-1 \\\\ & \Rightarrow f(1)=\frac{1}{4 \sqrt{2}}\left[\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right) \pi\right]-1 \end{aligned} $

Here, $f(1)<0$ and $f(0)=1>0$

$ \Rightarrow f(0) \cdot f(1)<0 $

By intermediate value theorem

$\Rightarrow$ There exists at least one $c \in(0,1)$ such that $f(c)=0$

Since, $f(x)$ is a decreasing function such that $f(x)=0$ has exactly one root.

$\alpha = \int\limits_0^1 {{e^{(9x + 3{{\tan }^{ - 1}}x)}}\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)dx} $

Set $9x + 3{\tan ^{ - 1}}x = t$

so that ${{dt} \over {dx}} = 9 + {3 \over {1 + {x^2}}} = {{12 + 9{x^2}} \over {1 + {x^2}}}$

We have, $\alpha = \int\limits_0^{9 + {{3\pi } \over 4}} {{e^t}dt = {e^{9 + {{3\pi } \over 4}}} - 1} $

$\therefore$ $\ln \left| {\alpha + 1} \right| = 9 + {{3\pi } \over 4}$

Thus $\ln \left| {\alpha + 1} \right| - {{3\pi } \over 4} = 9$

Given: $f: \mathrm{R} \rightarrow \mathrm{R} f(x)=\left\{\begin{aligned} {[x], } & x \leq 2 \\ o, & x>2\end{aligned}\right.$

$\text { And } \quad I=\int_\limits{-1}^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$

So, $f\left(x^2\right)=\left\{\begin{array}{cl} {\left[x^2\right],} & x^2 \leq 2, \quad x \in[-\sqrt{2}, \sqrt{2} \\ 0, & x^2>2, \quad x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty) \end{array}\right.$

$\text { And } f(x+1)=\left\{\begin{array}{cll} {[x+1],} & x+1 \leq 2, & x \leq 1 \\ 0, & x+1>2, & x>1 \end{array}\right.$

$\text { So, } \mathrm{I}=\int_\limits{-1}^0 \frac{x f\left[x^2\right]}{2+f(x+1)} d x+\int_\limits0^1 \frac{x f\left(x^2\right)}{2+f(x+1)} d x +\int_\limits1^{\sqrt{2}} \frac{x f\left(x^2\right)}{2+f(x+1)} d x+\int_\limits2^2 \frac{x f\left(x^2\right)}{2+f(x+1)} d x$

$\Rightarrow \mathrm{I}=\int_\limits{-1}^0 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits0^1 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits1^{\sqrt{2}} \frac{x\left[x^2\right]}{2+0} d x +\int_\limits{\sqrt{2}}^2 \frac{x \cdot 0}{2+0} d x$

$\Rightarrow \mathrm{I}=\int_\limits{-1}^0 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits0^1 \frac{x\left[x^2\right]}{2+[x+1]} d x+\int_\limits1^{\sqrt{2}} \frac{x\left[x^2\right]}{2} d x$

Using the property of greatest integer function, For $x \in(-1,0),[x+1]=0$ and $\left[x^2\right]=0$ and, For $x \in(0,1),[x+1]=1$ and $\left[x^2\right]=0$ and, For $x \in(1, \sqrt{2}),\left[x^2\right]=1$

$\begin{aligned} & \Rightarrow \quad \mathrm{I}=\int_{-1}^0 \frac{x .0}{2+0} d x+\int_0^1 \frac{x .0}{2+1} d x+\int_\limits1^{\sqrt{2}} \frac{x .1}{2} d x \\ & \Rightarrow \quad \mathrm{I}=\int_1^{\sqrt{2}} \frac{x}{2} d x \\ & \Rightarrow \quad \mathrm{I}=\frac{1}{2}\left[\frac{x^2}{2}\right]_1^{\sqrt{2}} \\ & \Rightarrow \quad \mathrm{I}=\frac{1}{2}\left[\frac{2}{2}-\frac{1}{2}\right] \\ & \Rightarrow \quad \mathrm{I}=\frac{1}{4} \\ & \Rightarrow \quad 4 \mathrm{I}=1 \\ & \Rightarrow 4 \mathrm{I}-1=0 \\ \end{aligned}$

Hint:

(i) Find $f\left(x^2\right)$ and $f(x+1)$ using composite function.

(ii) Split the given integral using the property of the greatest integer function.

(iii) Find the value of the definite integral using, if $\int g(x) d x=\mathrm{G}(x) \Rightarrow \int_a^b g(x) d x=[\mathrm{G}(b)-\mathrm{G}(a)]$

$\int\limits_0^1 \underbrace{4 x^3}_{\mathrm{I}} \underbrace{\frac{d^2}{d x^2}\left(1-x^2\right)^5}_{\mathrm{II}} d x$

Integrating by parts:

$ I=4 x^3\left[\frac{d}{d x}\left(1-x^2\right)^5\right]_0^1-\int\limits_0^1 12 x^2 \frac{d}{d x}\left(1-x^2\right)^5 d x $

$\begin{aligned} & =4 x^3\left[5\left(1-x^2\right)^4(-2 x)\right]_0^1-12\left[\left[x^2\left(1-x^2\right)^5\right]_0^1-\int\limits_0^1 2 x\left(1-x^2\right)^5 d x\right. \\\\ & =0-0+12 \int\limits_0^1 2 x\left(1-x^2\right)^5 d x\end{aligned}$

Now, putting $1-x^2=t$, we get $-2 x d r=d t$. Therefore,

$ I=-12 \int\limits_1^0 t^5 d t $

$\begin{aligned} & \text { When } x=0, t=1 . \\\\ & \text { When } x=1, t=0 .\end{aligned}$

$I=12 \times \int\limits_0^1 t^5 d t=12 \times\left[\frac{t^6}{6}\right]_0^1=12 \times \frac{1}{6}=2$

Case 1 :

Let $0 \le x < 1$

then $\left[ x \right] = 0$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 0 - x$

$ = 1 - x$

Case 2 :

Let $1 \le x < 2$

then $\left[ x \right] = 1$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 1$

Case 3 :

Let $2 \le x < 3$

then $\left[ x \right] = 2$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 2 - x$

$ = 3 - x$

Case 4 :

Let $3 \le x < 4$

then $\left[ x \right] = 3$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 3$

$\therefore$ $f(x) = \left\{ {\matrix{ {1 - x} & ; & {0 \le x < 1} \cr {x - 1} & ; & {1 \le x < 2} \cr {3 - x} & ; & {2 \le x < 3} \cr {x - 3} & ; & {3 \le x < 4} \cr } } \right.$

$\therefore$ $f(x)$ is periodic and period of $f(x) = 2$

And period of $\cos \pi x = {{2\pi } \over \pi } = 2$

$\therefore$ Period of $f(x)\cos \pi x = 2$

Now,

$I = {{{\pi ^2}} \over {10}}\int_{ - 10}^{10} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_{ - 10}^{ - 10 + 10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_0^{10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}} \times 10\int_0^2 {f(x)\cos \pi x\,dx} $

$ = {\pi ^2}\int_0^2 {f(x)\cos \pi x\,dx} $

$\therefore$ $I = {\pi ^2}\left[ {\int_0^1 {f(x)\cos \pi x\,dx + \int_1^2 {f(x)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {(1 - x)\cos \pi x\,dx + \int_1^2 {(x - 1)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {\cos \pi x\,dx - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - \int_1^2 {\cos \pi x\,dx} } } } } \right]$

$ = {\pi ^2}\left[ {{1 \over \pi }\left[ {\sin \pi x} \right]_0^1 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - {1 \over \pi }\left[ {\sin \pi x} \right]_1^2} } } \right]$

$ = {\pi ^2}\left[ {0 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - 0} } } \right]$

$ = {\pi ^2}\left[ { - \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_0^1 + \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_1^2} \right]$

$\left[ {\mathrm{As}\,\int {x\cos \pi x\,dx = x\,.\,\int {\cos \pi x - \int {\left( {1\,.\,{{\sin \pi x} \over \pi }} \right)dx = x\,.\,{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x + c} } } } \right]$

$ = {\pi ^2}\left[ { - \left[ {\left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\,.\,\cos \pi } \right) - \left( {0 + {1 \over {{\pi ^2}}}\,.\,\cos 0} \right)} \right] + \left[ {\left( {2\,.\,{{\sin 2\pi } \over \pi } + {1 \over {{\pi ^2}}}\cos 2\pi } \right) - \left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\cos \pi } \right)} \right]} \right]$

$ = {\pi ^2}\left[ { - \left\{ {\left( { - {1 \over {{\pi ^2}}}} \right) - \left( {{1 \over {{\pi ^2}}}} \right)} \right\} + \left( {\left( { + {1 \over {{\pi ^2}}}} \right) - \left( { - {1 \over {{\pi ^2}}}} \right)} \right.} \right]$

$ = {\pi ^2}\left[ { - \left( { - {2 \over {{\pi ^2}}}} \right) + {2 \over {{\pi ^2}}}} \right]$

$ = {\pi ^2}\left[ {{2 \over {{\pi ^2}}} + {2 \over {{\pi ^2}}}} \right]$

$ = {\pi ^2} \times {4 \over {{\pi ^2}}}$

$ = 4$

We have $f(x) = \int\limits_0^x {f(t)dt \Rightarrow f(0) = 0} $

Also, $f'(x) = f(x),x > 0$. Therefore, $f(x) = k,x > 0$

Hence, $f(0) = 0$ and $f(x)$ is continuous,

$f(x) = 0\forall x > 0$

Since $f(\ln 5) = 0$.

$ \begin{aligned} \mathrm{I} & =\frac{5050 \int_0^1\left(1-x^{50}\right)^{100} d x}{\int_0^1\left(1-x^{50}\right)^{100} d x} \\ & =5050 \frac{\mathrm{I}_{100}}{\mathrm{I}_{101}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Now, $\mathrm{I}_{101}=\int_0^1\left(1-x^{50}\right)^{101} d x$

Using integration by part

$ \begin{aligned} & \begin{aligned} = & {\left[\left(1-x^{50}\right)^{101} x\right]_0^1 } \end{aligned}+\int_0^1 101 \times 50\left(1-x^{50)^{100}} x^{49} x d x\right. \\ & =0+5050 \int_0^1 x^{50}\left(1-x^{50)^{100}} d x\right. \\ & =-5050 \int_0^1\left[\left(1-x^{50}\right)-1\right]\left(1-x^{50}\right)^{100} d x \\ & =-5050\left[\int_0^1\left(1-x^{50}\right)^{101} d x-\int_0^1\left(1-x^{50}\right)^{100} d x\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Using (ii) in (i), we have

$ \begin{aligned} \mathrm{I} & =\frac{5050 \mathrm{I}_{100}}{5050 \mathrm{I}_{100}} \times 5051 \\ & =5051 \end{aligned} $

$ \begin{array}{ll} & a_n=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\ldots(-1)^{n-1}\left(\frac{3}{4}\right)^n \\ & =\frac{3}{4}\left[\frac{1-\left(\frac{-3}{4}\right)^n}{1+\frac{3}{4}}\right]=\frac{3}{7}\left[1-\left(\frac{-3}{4}\right)^n\right] \\ & b_n>a_n \Rightarrow 2 a_n<1 \\ \Rightarrow & \frac{6}{7}\left[1-\left(\frac{-3}{4}\right)^n\right]<1 \\ \Rightarrow & 1-\left(\frac{-3}{4}\right)^n<\frac{7}{6} \\ \Rightarrow & \frac{-1}{6}<\left(\frac{-3}{4}\right)^n \\ \Rightarrow & \frac{1}{6}>\frac{-(-3)^n}{2^{2 n}} \\ \Rightarrow & 2^{2 n-1}>-(-3)^{n+1} \end{array} $

For $n$ to be even, inequality always holds.

For $n$ to be odd, it holds for $n \geq 7$.

Thus minimum natural number

$ n_0=6 $