Question

If $\mathrm{I}=\int \frac{dx}{{\left(2\sin x+\sec x\right)}^4}$. Then find the value of $\mathrm{I}$.

Answer 1

Step 1: Simplify the given integral

The given integral can be simplified as,

$\mathrm{I}=\int \frac{dx}{{\left(2\sin x+\sec x\right)}^4}$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\cos }^{4}x$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\cos }^{4}x$}\right.}}\times \frac{1}{{\left(2\sin x+\sec x\right)}^4}\right]dx$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\cos }^{4}x$}\right.}}{{\displaystyle \raisebox{1ex}{${\left(2\sin x+\sec x\right)}^4$}\!\left/ \!\raisebox{-1ex}{${\cos }^{4}x$}\right.}}\right]dx$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle {\sec }^{4}x}}{{\displaystyle {\left(\frac{{\displaystyle 2\sin x+\sec x}}{\cos x}\right)}^4}}\right]dx$ $\mathrm{[}\mathrm{\because }\mathrm{}\mathrm{secx}\mathrm{=}\frac{\mathrm{1}}{\mathrm{cosx}}\mathrm{]}$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle {\sec }^{4}x}}{{\displaystyle {\left(\frac{{\displaystyle 2\sin x}}{\cos x}+\frac{\sec x}{\cos x}\right)}^4}}\right]dx$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle {\sec }^{4}x}}{{\displaystyle {\left(2\tan x+\sec x·\sec x\right)}^4}}\right]dx$ $\mathrm{[}\mathrm{\because }\mathrm{}\mathrm{tanx}\mathrm{=}\frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{]}$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle {\sec }^{4}x}}{{\displaystyle {\left(2\tan x+{\sec }^{2}x\right)}^4}}\right]dx$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle {\sec }^{4}x}}{{\displaystyle {\left(2\tan x+1+{\tan }^{2}x\right)}^4}}\right]dx$ $\mathrm{[}\mathrm{\because }\mathrm{}{\sec }^{\mathrm{2}}\mathrm{x}\mathrm{=}\mathrm{1}\mathrm{+}{\tan }^{\mathrm{2}}\mathrm{x}\mathrm{]}$

$\Rightarrow \mathrm{I}=\int \left[\frac{{\displaystyle {\sec }^{4}x}}{{\displaystyle {\left\{{\left(1+\tan x\right)}^2\right\}}^4}}\right]dx$

$\Rightarrow \mathrm{I}=\int \frac{{\displaystyle {\sec }^{2}x·{\sec }^{2}x}}{{\displaystyle {\left(1+\tan x\right)}^8}}dx$ …(i)

Step 2: Assume and deduce values

Now, let, $1+\tan x=t$ …(ii)

$\Rightarrow$ $\tan x=t-1$

$\Rightarrow$ ${\tan }^{2}x={\left(t-1\right)}^2$ [Squaring on both sides]

$\Rightarrow$ ${\sec }^{2}x-1={\left(t-1\right)}^2$ $\mathrm{[}\mathrm{\because }\mathrm{}{\sec }^{\mathrm{2}}\mathrm{x}\mathrm{=}\mathrm{1}\mathrm{+}{\tan }^{\mathrm{2}}\mathrm{x}\mathrm{]}$

$\Rightarrow$ ${\sec }^{2}x={\left(t-1\right)}^2+1$ …(iii)

Differentiating equation (ii) with respect to $x$,

$\frac{d}{dx}1+\frac{d}{dx}\tan x=\frac{dt}{dx}$

$\Rightarrow$ $0+\frac{1}{{\cos }^{2}x}=\frac{dt}{dx}$ $\mathrm{[}\mathrm{\because }\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{C}\mathrm{=}\mathrm{0}\mathrm{,}\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{tanx}\mathrm{=}\frac{\mathrm{1}}{{\cos }^{\mathrm{2}}\mathrm{x}}\mathrm{]}$

$\Rightarrow$ ${\sec }^{2}x=\frac{dt}{dx}$ $\mathrm{[}\mathrm{\because }\mathrm{}\mathrm{secx}\mathrm{=}\frac{\mathrm{1}}{\mathrm{cosx}}\mathrm{]}$

$\Rightarrow$ ${\sec }^{2}xdx=dt$ …(iv)

Step 3: Substitute values and integrate

Now, substituting the values obtained in equations (ii), (iii) and (iv) into the equation (i), we get,

$\mathrm{I}=\int \frac{{\displaystyle {\left(t-1\right)}^2+1}}{{\displaystyle {\left(t\right)}^8}}dt$

$\Rightarrow \mathrm{I}=\int \frac{{\displaystyle t^2+1-2t+1}}{{\displaystyle t^8}}dt$ $\mathrm{[}\mathrm{\because }\mathrm{}{\left(a-b\right)}^{\mathrm{2}}\mathrm{=}{\mathrm{a}}^{\mathrm{2}}\mathrm{+}{\mathrm{b}}^{\mathrm{2}}\mathrm{-}\mathrm{2}\mathrm{ab}\mathrm{]}$

$\Rightarrow \mathrm{I}=\int \frac{{\displaystyle t^2-2t+2}}{{\displaystyle t^8}}dt$

$\Rightarrow \mathrm{I}=\int \left(\frac{{\displaystyle t^2}}{{\displaystyle t^8}}-\frac{2t}{t^8}+\frac{2}{t^8}\right)dt$

$\Rightarrow \mathrm{I}=\int \left(t^{2-8}-2t^{1-8}+2t^{-8}\right)dt$ $\mathrm{[}\mathrm{\because }\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}\mathrm{=}{\mathrm{a}}^{\mathrm{m}\mathrm{-}\mathrm{n}}\mathrm{]}$

$\Rightarrow \mathrm{I}=\int \left(t^{-6}-2t^{-7}+2t^{-8}\right)dt$

$\Rightarrow \mathrm{I}=\int t^{-6}dt-\int 2t^{-7}dt+\int 2t^{-8}dt$

$\Rightarrow \mathrm{I}=\frac{t^{-6+1}}{-6+1}-2\frac{t^{-7+1}}{-7+1}+2\frac{t^{-8+1}}{-8+1}+C$ $\mathrm{[}\mathrm{\because }\mathrm{\int }{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}\mathrm{=}\frac{{\mathrm{x}}^{\mathrm{n}\mathrm{+}\mathrm{1}}}{\mathrm{n}\mathrm{+}\mathrm{1}}\mathrm{+}\mathrm{C}\mathrm{]}$

$\Rightarrow \mathrm{I}=\frac{t^{-5}}{-5}-2\frac{t^{-6}}{-6}+2\frac{t^{-7}}{-7}+C$

$\Rightarrow \mathrm{I}=-\frac{1}{5t^5}+\frac{2}{6t^6}-\frac{2}{7t^7}+C$

$\Rightarrow \mathrm{I}=-\frac{1}{5t^5}+\frac{1}{3t^6}-\frac{2}{7t^7}+C$

Substituting $t=1+\tan x$, we get,

$\mathrm{I}=-\frac{1}{5{\left(1+\tan x\right)}^5}+\frac{1}{3{\left(1+\tan x\right)}^6}-\frac{2}{7{\left(1+\tan x\right)}^7}+C$

Hence, the value of $\mathrm{I}$ is $-\frac{1}{5{\left(1+\tan x\right)}^5}+\frac{1}{3{\left(1+\tan x\right)}^6}-\frac{2}{7{\left(1+\tan x\right)}^7}+C$.