Question

The integral $\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}{e}^x\left(\ln \right(\cos x\left)\right)(\cos x\sin x-1){\text{cosec}}^{2}xdx$ is equal to

• A. $\frac{e^{\pi /4}}{2}[2e^{\pi /4}+\ln 2-2]$
• B. $\frac{e^{\pi /4}}{2}[e^{\pi /4}+\ln 2-2]$
• C. $\frac{e^{\pi /4}}{2}[e^{\pi /4}-\ln 2+2]$
• D. $\frac{e^{\pi /4}}{2}[2e^{\pi /4}-\ln 2+2]$

Answer 1

The correct option is A $\frac{e^{\pi /4}}{2}[2e^{\pi /4}+\ln 2-2]$

$\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}{e}^x\left(\ln \right(\cos x\left)\right)(\cos x\sin x-1){\text{cosec}}^{2}xdx$
Let
$I=\int \left(\ln \right(\cos x\left)\right)\cdot e^x(\cot x-{\text{cosec}}^{2}x)dx$
We know that,
$\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)=e^{x}f\left(x\right)+c\Rightarrow \int e^x(\cot x-{\text{cosec}}^{2}x)dx=e^x\cot x+c$
And using by parts, we get
$\Rightarrow I=\ln \left(\cos x\right)e^x\cot x+\int \tan x\cdot e^x\cdot \cot xdx\Rightarrow I=\ln \left(\cos x\right)e^x\cot x+\int e^{x}dx+c\Rightarrow I=e^x\left[\ln \right(\cos x)\cdot \cot x+1]+c$
So,
$\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}{e}^x\left(\ln \right(\cos x\left)\right)(\cos x\sin x-1){\text{cosec}}^{2}xdx={\left(e^x\right[\ln \left(\cos x\right)\cdot \cot x+1\left]\right)}_{\pi /4}^{\pi /2}=e^{\pi /2}\left[\underset{x\to \pi /2}{lim}\ln \right(\cos x)\cdot \cot x]+e^{\pi /2}-e^{\pi /4}[1-\ln \sqrt{2}]$
Now, solving the limit we get
$\underset{x\to \pi /2}{lim}\ln \left(\cos x\right)\cdot \cot x=\underset{x\to \pi /2}{lim}\frac{\ln \left(\cos x\right)}{\tan x}$
Using L'Hospital's Rule, we get
$=\underset{x\to \pi /2}{lim}\frac{-\sin x}{\cos x\cdot {\sec }^{2}x}=\underset{x\to \pi /2}{lim}-\sin x\cos x=0$

Therefore,
$e^{\pi /2}\left[\underset{x\to \pi /2}{lim}\ln \right(\cos x)\cdot \cot x]+e^{\pi /2}-e^{\pi /4}[1-\ln \sqrt{2}]=e^{\pi /2}-e^{\pi /4}[1-\ln \sqrt{2}]=\frac{e^{\pi /4}}{2}[2e^{\pi /4}+\ln 2-2]$