Indefinite Integrals

$ \begin{aligned} & \int(\sin x)^{\frac{-11}{2}}(\cos x)^{\frac{-5}{2}} d x=\int \frac{1}{\operatorname{Tan}^{11 / 2} x} \cdot \sec ^8 x d x=\int \frac{\left(1+\operatorname{Tan}^2 x\right)^3 \sec ^2 x}{\operatorname{Tan}^{11 / 2} x} d x \\ & \text { put } \operatorname{Tan} x=t \\ & \frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}=\frac{15(2 \times 6 \times 6 \times 2)}{9 \times 5 \times 1 \times 3}=16 \end{aligned} $

$\begin{aligned} & \int\left(\frac{1}{x^2}+\frac{1}{x^4}\right)\left(\frac{3}{x}+\frac{1}{x^3}\right)^{\frac{1}{23}} d x \\ & \text { using } t=\frac{3}{x}+\frac{1}{x^3} \Rightarrow d t=-3\left(\frac{1}{x^2}+\frac{1}{x^4}\right) d x \\ & \int \frac{t^{1 / 23} d t}{-3}=\frac{t^{24 / 23}}{\left(\frac{24}{23}\right)(-3)}+C \\ & \Rightarrow \alpha=23 \beta=-1 \gamma=-3 \\ & \alpha+\beta+\gamma=19 \end{aligned}$

$\begin{aligned} & \sqrt{1+x^2}+x=\sec \theta+\tan \theta=t \\ & \sqrt{1+x^2}=\sec \theta=\frac{t^2+1}{2 t} \\ & x=\tan \theta=\frac{t^2-1}{2 t} \end{aligned}$

The given expression becomes

$\frac{1}{m} t^2\left(n \cdot \frac{t^2+1}{2 t}-\frac{t^2-1}{2 t}\right)=\frac{t^{n-1}}{2 m}\left((n-1) t^2+n+1\right)$

By compare

$\begin{aligned} & n=19 \\ & m=360 \\ & \therefore \quad n+m=379 \end{aligned}$

$\begin{aligned} & 2 \mathrm{x}^2+5 \mathrm{x}+9=\mathrm{A}\left(\mathrm{x}^2+\mathrm{x}+1\right)+\mathrm{B}(2 \mathrm{x}+1)+\mathrm{C} \\ & \mathrm{~A}=2 \quad \mathrm{~B}=\frac{3}{2} \quad \mathrm{C}=\frac{11}{2} \end{aligned}$

$\begin{aligned} & 2 \int \sqrt{x^2+x+1} d x+\frac{3}{2} \int \frac{2 x+1}{\sqrt{x^2+x+1}} d x+\frac{11}{2} \int \frac{d x}{\sqrt{x^2+x+1}} \\ & 2 \int \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \mathrm{dx}+3 \sqrt{x^2+x+1}+\frac{11}{2} \int \frac{\mathrm{dx}}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}} \\ & 2\left(\frac{x+\frac{1}{2}}{2} \sqrt{x^2+x+1}+\frac{3}{8} \ell n\left(x+\frac{1}{2}+\sqrt{x^2+x+1}\right)\right)+3 \sqrt{x^2+x+1} \\ & +\frac{11}{2} \ell n\left(x+\frac{1}{2}+\sqrt{x^2+x+1}\right)+C \\ & \alpha=\frac{7}{2} \\ & \alpha+2 \beta=16 \end{aligned}$

$\begin{aligned} & I=\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} d x \\ & \frac{x+3}{x-1}=t \Rightarrow d x=\frac{-4}{(t-1)^2} d t \Rightarrow x=\left(\frac{3+t}{t-1}\right) \\ & \Rightarrow(x-1)^4(x+3)^6=(x-1)^5(x+3)^5\left(\frac{x+3}{x-1}\right) \\ & I=\int \frac{\frac{-4}{(t-1)^2} d t}{t^{1 / 5}\left(\frac{3+t}{t-1}-1\right)\left(\frac{3+t}{t-3}+3\right)} \\ & I=\int \frac{-4 d t}{t^{1 / 5}(16 t)}=\frac{5}{4}\left(\frac{x-1}{x+3}\right)^{1 / 5}+c \end{aligned}$

Comparing,

$\begin{aligned} & \Rightarrow A=\frac{5}{4}, B=\frac{1}{5}, \alpha=1, \beta=1 \\ & \Rightarrow \alpha+\beta+20 A B=1+1+20 \times \frac{5}{4} \times \frac{1}{5}=7 \end{aligned}$

$\begin{aligned} & I=\int(\operatorname{cosec} x)^5 d x=\int(\operatorname{cosec} x)^3(\operatorname{cosec} x)^2 d x \\ & =(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x- \\ & \int\left(\frac{d}{d x}(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x\right) d x \end{aligned}$

$\begin{aligned} & =-\cot x(\operatorname{cosec} x)^3-\int 3 \operatorname{cosec}^2 x \cdot(-\operatorname{cosec} x \cot x)(-\cot x) d x \\ & =-\cot x(\operatorname{cosec} x)^3-\int 3 \operatorname{cosec}^3 x \cot ^2 x d x \\ & =-\cot x(\operatorname{cosec} x)^3-3 \int(\operatorname{cosec} x)^3\left(\operatorname{cosec}^2 x-1\right) d x \\ & I=-\cot x(\operatorname{cosec} x)^3-3 I+3 \int(\operatorname{cosec} x)^3 d x \\ & 4 I=-\cot x(\operatorname{cosec} x)^3+3 \int(\operatorname{cosec} x)^3 d x \\ & =-\cot x(\operatorname{cosec} x)^3+3 I_1 \\ & I_1=\int \operatorname{cosec} x \cdot \operatorname{cosec}^2 x d x=\operatorname{cosec} x(-\cot x)- \\ & \qquad \int(-\operatorname{cosec} x \cot x)(-\cot x) d x \end{aligned}$

$\begin{aligned} & I_1=-\operatorname{cosec} x \cot x-\int \operatorname{cosec} x\left(\operatorname{cosec}^2 x-1\right) d x \\ & I_1=-\operatorname{cosec} x \cot x-I+\int \operatorname{cosec} x d x \\ & 2 l_1=-\operatorname{cosec} x \cot x+\ln \left|\tan \frac{x}{2}\right| \\ & \Rightarrow \quad 4 l=-\cot x(\operatorname{cosec} x)^3-\frac{3}{2} \operatorname{cosec} x \cot x +\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+C \end{aligned}$

$\begin{aligned} & I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+C \\ & \Rightarrow \alpha=-\frac{1}{4}, \beta=\frac{3}{8} \Rightarrow 8(\alpha+\beta)=1 \end{aligned}$

$ \begin{aligned} & f(x)=\int \frac{d x}{\left(3+4 x^2\right) \sqrt{4-3 x^2}} \\\\ & x=\frac{1}{t} \\\\ & =\int \frac{\frac{-1}{t^2} d t}{\frac{\left(3 t^2+4\right)}{t^2} \frac{\sqrt{4 t^2-3}}{t}} \\\\ & =\int \frac{-d t \cdot t}{\left(3 t^2+4\right) \sqrt{4 t^2-3}}: \text { Put } 4 t^2-3=z^2 \\\\ & =-\frac{1}{4} \int \frac{z d z}{\left(3\left(\frac{z^2+3}{4}\right)+4\right) z} \end{aligned} $

$ \begin{aligned} & =\int \frac{-d z}{3 z^2+25}=-\frac{1}{3} \int \frac{\mathrm{dz}}{\mathrm{z}^2+\left(\frac{5}{\sqrt{3}}\right)^2} \\\\ & =-\frac{1}{3} \frac{\sqrt{3}}{5} \tan ^{-1}\left(\frac{\sqrt{3} z}{5}\right)+C \\\\ & =-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{4 t^2-3}\right)+C \\\\ & f(x)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{\frac{4-3 \mathrm{x}^2}{x^2}}\right)+C \\\\ & \because \mathrm{f}(0)=0 \because \mathrm{c}=\frac{\pi}{10 \sqrt{3}} \end{aligned} $

$ \begin{aligned} & f(1)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5}\right)+\frac{\pi}{10 \sqrt{3}} \\\\ & f(1)=\frac{1}{5 \sqrt{3}} \cot ^{-1}\left(\frac{\sqrt{3}}{5}\right)=\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{5}{\sqrt{3}}\right) \\\\ & \alpha=5: \beta=\sqrt{3} \therefore \alpha^2+\beta^2=28 \end{aligned} $

Given integral: $\int \sqrt{\frac{x+7}{x}} \, dx$

Let's make the substitution $x = t^2$. Then, $dx = 2t \, dt$.

Substituting these values, the integral becomes :

$\int 2 \sqrt{t^2 + 7} \, dt$

Now, let's evaluate this integral :

$I(t) = 2\left(\frac{t}{2} \sqrt{t^2+7} + \frac{7}{2} \ln\left|t + \sqrt{t^2+7}\right|\right) + C$

Substituting back $t = \sqrt{x}$, we have :

$I(x) = 2\left(\frac{\sqrt{x}}{2} \sqrt{x+7} + \frac{7}{2} \ln\left|\sqrt{x} + \sqrt{x+7}\right|\right) + C$

Simplifying further :

$I(x) = \sqrt{x} \sqrt{x+7} + 7 \ln\left|\sqrt{x} + \sqrt{x+7}\right| + C$

We are given that $I(9) = 12+7 \ln 7$.

Let's substitute $x = 9$ and solve for the constant $C$:

$12+7 \ln 7 = \sqrt{9} \sqrt{9+7}+7 \ln |\sqrt{9}+\sqrt{9+7}|+C$

$ \Rightarrow $ $12+7 \ln 7 = 3 \sqrt{16}+7 \ln |\sqrt{9}+\sqrt{16}|+C$

$ \Rightarrow $ $12+7 \ln 7 = 3 \cdot 4+7 \ln (3+\sqrt{16})+C$

$ \Rightarrow $ $12+7 \ln 7 = 12+7 \ln (3+4)+C$

$ \Rightarrow $ $12+7 \ln 7 = 12+7 \ln 7+C$

From this equation, we can see that $C = 0$.

Now, we need to calculate $I(1)$ :

$I(1) = \sqrt{1} \sqrt{1+7}+7 \ln |\sqrt{1}+\sqrt{1+7}|$

$ \Rightarrow $ $I(1) = 1 \sqrt{8}+7 \ln (1+\sqrt{8})$

$ \Rightarrow $ $I(1) = \sqrt{8}+7 \ln (1+2 \sqrt{2})$

Therefore, $\alpha = \sqrt{8}$.

Finally, to find $\alpha^4$:

$\alpha^4 = \left(\sqrt{8}\right)^4$

$ \Rightarrow $ $\alpha^4 = 8^2$

$ \Rightarrow $ $\alpha^4 = 64$

Hence, $\alpha^4$ is equal to 64.

$I = \int {\sqrt {\sec 2x - 1} dx=\int {\sqrt {{{1 - \cos 2x} \over {\cos 2x}}} dx} } $

$ = \int {\sqrt {{{2{{\sin }^2}x} \over {2{{\cos }^2}x - 1}}} dx} $

Let $\sqrt {2\cos } x = t - \sqrt 2 \sin xdx = dt$

$I = \int { - {{dt} \over {\sqrt {{t^2} - 1} }} = - \ln \left| {t + \sqrt {{t^2} - {a^2}} } \right| + c} $

$ = - \ln \left| {\sqrt 2 \cos x + \sqrt {2{{\cos }^2}} x - 1} \right| + c$

$ = - {1 \over 2}\ln \left| {2{{\cos }^2}x + \cos 2x + 2\sqrt 2 \sqrt {\cos 2x.{{\cos }^2}x} } \right| + c$

$ = - {1 \over 2}\ln \left| {2\cos 2x + 1 + 2\sqrt {\cos 2x(1 + \cos 2x)} } \right| + c$

$ = - {1 \over 2}\ln \left| {\cos 2x + {1 \over 2} + \sqrt {\cos 2x(1 + \cos 2x)} } \right| + c$

$\therefore$ $\alpha = {{ - 1} \over 2},\beta = {1 \over 2}$

$\therefore$ $\beta - \alpha = 1$

$ = \int {{{{{\sin x} \over {{{\cos }^3}x}}} \over {1 + {{\tan }^3}x}}dx = \int {{{\tan x.{{\sec }^2}x} \over {(\tan x + 1)(1 + {{\tan }^2}x - \tan x)}}dx} } $

Let $\tan x = t \Rightarrow {\sec ^2}x.\,dx = dt$

$ = \int {{t \over {(t + 1)({t^2} - t + 1)}}dt} $

$ = \int {\left( {{A \over {t + 1}} + {{B(2t - 1)} \over {{t^2} - t + 1}} + {C \over {{t^2} - t + 1}}} \right)dx} $

$ \Rightarrow A({t^2} - t + 1) + B(2t - 1)({t^2} - t + 1) + C(t + 1) = t$

$ \Rightarrow {t^2}(A + 2B) + t( - A + B + C) + A - B + C = 1$

$\therefore$ $A + 2B = 0$ ..... (1)

$ - A + B + C = 1$ ... (2)

$A - B + C = 0$ ... (3)

$ \Rightarrow C = {1 \over 2} \Rightarrow A - B = - {1 \over 2}$ ... (4)

$A + 2B = 0$

$A - B = - {1 \over 2}$

$\Rightarrow$ $3B = {1 \over 2} \Rightarrow B = {1 \over 6}$

$A = - {1 \over 3}$

$I = - {1 \over 3}\int {{{dt} \over {1 + t}} + {1 \over 6}\int {{{2t - 1} \over {{t^2} - t + 1}}dt + {1 \over 2}\int {{{dt} \over {{t^2} - t + 1}}} } } $

$ = - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over 2}.{2 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{\left( {\tan x - {1 \over 2}} \right)} \over {{{\sqrt 3 } \over 2}}}} \right)$

$ = - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C$

$\alpha = - {1 \over 3},\beta = {1 \over 6},\gamma = {1 \over {\sqrt 3 }}$

$18(\alpha + \beta + {\gamma ^2}) = 18\left( { - {1 \over 3} + {1 \over 6} + {1 \over 3}} \right) = 3$

$2{e^x} + 3{e^{ - x}} = A(4{e^x} + 7{e^{ - x}}) + B(4{e^x} - 7{e^{ - x}}) + \lambda $

2 = 4A + 4B ; 3 = 7A $-$ 7B ; $\lambda$ = 0

$A + B = {1 \over 2}$

$A - B = {3 \over 7}$

$A = {1 \over 2}\left( {{1 \over 2} + {3 \over 7}} \right) = {{7 + 6} \over {28}} = {{13} \over {28}}$

$B = A - {3 \over 7} = {{13} \over {28}} - {3 \over 7} = {{13 - 12} \over {28}} = {1 \over {28}}$

$\int {{{13} \over {28}}dx + {1 \over {28}}\int {{{4{e^x} - 7{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}} dx} $

= ${{13} \over {28}}x + {1 \over {28}}\ln |4{e^x} + 7{e^{ - x}}| + C$

$u = {{13} \over 2};v = {1 \over 2}$

$\Rightarrow$ u + v = 7

$\int \frac{d x}{\left(x^2+x+1\right)^2}$

$=\int \frac{d x}{\left[\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]^2}$

Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2} \tan \theta$

$ \Rightarrow d x=\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta $

$\therefore \int \frac{\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta}{\frac{9}{16}\left(\tan ^2 \theta+1\right)^2} $

$ =\frac{8}{3 \sqrt{3}} \int \frac{\sec ^2 \theta d \theta}{\sec ^4 \theta} $

$ =\frac{8}{3 \sqrt{3}} \int \cos ^2 \theta d \theta=\frac{8}{3 \sqrt{3}} \int \frac{1+\cos 2 \theta}{2} d \theta $

$ =\frac{4}{3 \sqrt{3}}\left(\theta+\frac{\sin 2 \theta}{2}\right)+C $

= $ \frac{4}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+\frac{4}{3 \sqrt{3}} \frac{\frac{2 x+1}{\sqrt{3}}}{1+\left(\frac{2 x+1}{\sqrt{3}}\right)^2}+C$

$=\frac{4}{3 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{3}\right)+\frac{1}{3} \frac{2 x+1}{\left(x^2+x+1\right)}+C$

$\therefore a=\frac{4}{3 \sqrt{3}}, b=\frac{1}{3}$

Hence, $9(\sqrt{3} a+b)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$

$\int {{{5{x^8} + 7{x^6}} \over {{{(2{x^7} + {x^2} + 1)}^2}}}dx = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} } $

$\int {{{{5 \over {{x^6}}} + {7 \over {{x^8}}}} \over {{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} $

put $2 + {1 \over {{x^5}}} + {1 \over {{x^7}}} = t$

$ \Rightarrow - \left( {{5 \over {{x^6}}} + {7 \over {{x^8}}}} \right)dx = dt$

$\int {{{ - dt} \over {{t^2}}} = {1 \over t} + c} $

$ \Rightarrow f(x) = {1 \over {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}}} + C = {{{x^7}} \over {2{x^7} + 1 + {x^2}}} + C$

$f(0) = 0 \Rightarrow C = 0$

$f(x) = {1 \over 4} = {1 \over k}$

$ \Rightarrow k = 4$

$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $

= $\int {{{{x^2} - 1} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx + \int {{1 \over {{x^4} + 3{x^2} + 1}}dx} } $

Let, ${I_1} = \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)}}} dx$

and ${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $

${\tan ^{ - 1}}\left( {x + {1 \over x}} \right) = t$

${I_1} = \int {{{dt} \over t}} $

${I_1} = \ln (t) = \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right|$

Now

${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $

$ = {1 \over 2}\int {{{({x^2} + 1) - ({x^2} - 1)} \over {{x^4} + 3{x^2} + 1}}} dx$

$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx - } \int {{{\left( {1 - {1 \over {{x^2}}}} \right)} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx} } \right]$

$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {\left[ {{{\left( {x - {1 \over x}} \right)}^2} + 5} \right]}}} dx - \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]}}} dx} \right]$

Let ${x - {1 \over x} = u}$ and ${x + {1 \over x} = v}$

$ = {1 \over 2}\left[ {\int {{{du} \over {{u^2} + {{\left( {\sqrt 5 } \right)}^2}}} - \int {{{dv} \over {{v^2} + 1}}} } } \right]$

${I_2} = {1 \over {2\sqrt 5 }}{\tan ^{ - 1}}\left( {{{x - {1 \over x}} \over {\sqrt 5 }}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {x + {1 \over x}} \right)$

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

$ \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right| + {1 \over {2\sqrt 5 }}\ln \left( {{{{x^2} - 1} \over {\sqrt 5 x}}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C$

$\alpha = 1,\beta = {1 \over {2\sqrt 5 }},\lambda = {1 \over {\sqrt 5 }},\delta = - {1 \over 2}$

$ \therefore $ $10(\alpha + \beta \lambda + \delta ) = 10\left[ {1 + {1 \over {10}} - {1 \over 2}} \right]$

$ = 10\left( {{1 \over {10}} + {1 \over 2}} \right)$

$ = 1 + 5 = 6$