Question

Evaluate :$\int \frac{\cos x}{{\sin }^{2}x(\sin x+\cos x)}dx$

• A. $\log \left|\frac{(1+\tan x)}{\tan x}\right|-\cot \left(x\right)+C$
• B. $\log \left|\frac{t\mathrm{an}\left(x\right)}{1+\tan \left(x\right)}\right|+C$
• C. $\log \left|\frac{t\mathrm{an}\left(x\right)}{1+\tan \left(x\right)}\right|-\tan \left(x\right)+C$
• D. $\log \left|\frac{\tan \left(x\right)}{(1+\tan x)}\right|+\cot \left(x\right)+C$

Answer 1

The correct option is A

$\log \left|\frac{(1+\tan x)}{\tan x}\right|-\cot \left(x\right)+C$

Explanation for the correct option:

Step 1: Using the substitution method,

Divide numerator and denominator by ${\cos }^3\left(x\right)$

$$\begin{gathered} I=\int \frac{{\displaystyle \frac{\cos x}{{\cos }^3\left(x\right)}}}{{\displaystyle \frac{{\sin }^{2}x(\sin x+\cos x)}{{\cos }^3\left(x\right)}}}dx \\ =\int \frac{{\displaystyle \frac{1}{{\cos }^2\left(x\right)}}}{{\displaystyle \frac{{\tan }^2\left(x\right)(\sin \left(x\right)+\cos \left(x\right))}{\cos \left(x\right)}}}dx \\ =\int \frac{{\displaystyle {\sec }^2\left(x\right)}}{{\displaystyle {\tan }^2\left(x\right)(\tan \left(x\right)+1)}}dx \end{gathered}$$

Let $t=\tan \left(x\right)\Rightarrow dt={\sec }^2\left(x\right)dx$

$$\begin{gathered} =\int \frac{{\displaystyle dt}}{{\displaystyle t^2(t+1)}} \\ =\int \frac{{\displaystyle dt}}{{\displaystyle t^2(t+1)}} \end{gathered}$$

Step 2: Integration of Partial fraction:

$$\begin{gathered} \frac{1}{t^2\left(t+1\right)}=\frac{A}{t}+\frac{B}{t^2}+\frac{C}{\left(t+1\right)} \\ equatingthenumerator \\ 1=At\left(t+1\right)+B\left(t+1\right)+C{t}^2 \\ Lett=-1 \\ \Rightarrow C=1 \\ Lett=0 \\ \Rightarrow B=1 \\ Lett=1 \\ 1=2A+2B+C \\ 1=2A+2\times 1+1 \\ 1-3=2A \\ \Rightarrow A=\frac{-2}{2}=-1 \end{gathered}$$

Step 3: substitute these values in integral

$$\begin{gathered} \int \frac{{\displaystyle dt}}{{\displaystyle t^2(t+1)}}=\int \left[-\frac{1}{t}+\frac{1}{t^2}+\frac{1}{\left(t+1\right)}\right]dt \\ =-\log \left|t\right|-\frac{1}{t}+\log \left|t+1\right|+C \\ =-\log \left|\tan \left(x\right)\right|-\frac{1}{\tan \left(x\right)}+\log \left|\tan \left(x\right)+1\right|+C \\ =-\cot \left(x\right)+\log \left|\frac{\tan \left(x\right)+1}{\tan \left(x\right)}\right|+C\left[\because \log \frac{a}{b}=\log a-\log b\right] \end{gathered}$$

Hence, option (A) is the correct answer.