Question

Solve $\int e^x\left(\frac{2\tan x}{1+\tan x}+{\cot }^2\left(x+\frac{\pi }{4}\right)\right)dx$

• A. $e^x\tan \left(\frac{\pi }{4}-x\right)+c$
• B. $e^x\tan \left(x-\frac{\pi }{4}\right)+c$
• C. $e^x\tan \left(\frac{3\pi }{4}-x\right)+c$
• D. None of these

Answer 1

Answer: (B)

$e^x\tan \left(x-\frac{\pi }{4}\right)+c$

consider the given function

$I=\int e^x(\frac{2\tan x}{1+\tan x}+{\cot }^2(x+\frac{\pi }{4}))dx$

First we solve that

${\cot }^2(x+\frac{\pi }{4})=\frac{1}{ta{n}^2(x+\frac{\pi }{4})}$

Then,

$\tan (x+\frac{\pi }{4})=\frac{\tan x+\tan \frac{\pi }{4}}{1-\tan x\tan \frac{\pi }{4}}$

$=\frac{1+\tan x}{1-\tan x}$

On reciprocal that,

$\frac{1}{\tan (x+\frac{\pi }{4})}=\frac{1-\tan x}{1+\tan x}$

$\cot (x+\frac{\pi }{4})=\frac{1-\tan x}{1+\tan x}$

Squaring both side and we get,

${\cot }^2(x+\frac{\pi }{4})={\left(\frac{1-\tan x}{1+\tan x}\right)}^2$

${\cot }^2(x+\frac{\pi }{4})={\left(\tan (x-\frac{\pi }{4})\right)}^2$

We know that,

${\tan }^2\theta ={\sec }^2\theta -1$

Then,

${\tan }^2(x-\frac{\pi }{4})={\sec }^2(x-\frac{\pi }{4})-1$

Put the value of given function

$I=\int e^x(\frac{2\tan x}{1+\tan x}+{\sec }^2(x-\frac{\pi }{4})-1)dx$

$I=\int e^x(\frac{2\tan x}{1+\tan x}-1+{\sec }^2(x-\frac{\pi }{4}))dx$

$I=\int e^x(\frac{2\tan x-1-\tan x}{1+\tan x}+{\sec }^2(x-\frac{\pi }{4}))dx$

$I=\int e^x(\frac{\tan x-1}{1+\tan x}+{\sec }^2(x-\frac{\pi }{4}))dx$

$I=\int e^x(\tan (x-\frac{\pi }{4})+{\sec }^2(x-\frac{\pi }{4}))dx$

But we know that,

$I=\int e^x(f\left(x\right)+f^{\prime }\left(x\right))dx$

$\int e^x(f\left(x\right)+f^{\prime }\left(x\right))dx=e^{x}f\left(x\right)+C$

Then,

$I=\int e^x(\tan (x-\frac{\pi }{4})+{\sec }^2(x-\frac{\pi }{4}))dx$

$\int e^x(\tan (x-\frac{\pi }{4})+{\sec }^2(x-\frac{\pi }{4}))dx=\int e^x\tan (x-\frac{\pi }{4})dx+C$

Hence, this is the answer.