Q104 / 200 · Manipal (MET) 2014 · Mathematics
$\cdot \int \frac{\cos x}{\sin ^2 x \cdot(\sin x+\cos x)} d x$ is equal to
- Correct answerA$\log \left|\frac{1+\tan x}{\tan x}\right|-\cot x+C$
- B$\log \left|\frac{\tan x}{1+\tan x}\right|+C$
- C$\log \left|\frac{\tan x}{1+\tan x}\right|-\tan x+C$
- D$\log \left|\frac{\tan x}{1+\tan x}\right|+\cot x+C$
Solution
Let, $I=\int \frac{\cos x}{\sin ^2 x(\sin x+\cos x)} d x$
On dividing numerator and denominator by $\cos ^3 x$, we get
$I=\int \frac{\sec ^2 x}{\tan ^2 x(1+\tan x)} d x$
Put $\quad \tan x=t \Rightarrow \sec ^2 x d x=d t$ $\begin{aligned} \therefore \quad I & =\int \frac{d t}{t^2(1+t)} \\ & =\int\left[-\frac{1}{t}+\frac{1}{t^2}+\frac{1}{1+t}\right] d t \end{aligned}$ (using partial fraction) $\begin{aligned} \Rightarrow \quad I & =-\int \frac{1}{t} d t+\int t^{-2} d t+\int \frac{1}{1+t} d t \\ & =-\log |t|-\frac{1}{t}+\log |1+t|+C \\ & =-\frac{1}{\tan x}+\log \left|\frac{1+\tan x}{\tan x}\right|+C \\ & =-\cot x+\log \left|\frac{1+\tan x}{\tan x}\right|+C \end{aligned}$
Put $\quad \tan x=t \Rightarrow \sec ^2 x d x=d t$ $\begin{aligned} \therefore \quad I & =\int \frac{d t}{t^2(1+t)} \\ & =\int\left[-\frac{1}{t}+\frac{1}{t^2}+\frac{1}{1+t}\right] d t \end{aligned}$ (using partial fraction) $\begin{aligned} \Rightarrow \quad I & =-\int \frac{1}{t} d t+\int t^{-2} d t+\int \frac{1}{1+t} d t \\ & =-\log |t|-\frac{1}{t}+\log |1+t|+C \\ & =-\frac{1}{\tan x}+\log \left|\frac{1+\tan x}{\tan x}\right|+C \\ & =-\cot x+\log \left|\frac{1+\tan x}{\tan x}\right|+C \end{aligned}$
Hence, the correct answer is option: 'A'.