Question

$\int e^x(\frac{2\tan x}{1+\tan x}+{\cot }^2(x+\frac{\pi }{4}))dx$ is equal to

• A. $e^x\tan (\frac{\pi }{4}-x)+c$
• B. $e^x\tan (x-\frac{\pi }{4})+c$
• C. $e^x\tan (\frac{3\pi }{4}-x)+c$
• D. $e^x\tan (x-\frac{3\pi }{8})+c$

Answer 1

The correct option is B

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

$\int e^x[\frac{2}{1+\cot x}-1+cose{c}^2(x+\frac{\pi }{4})]dx$

$=\int e^x[-\cot (x+\frac{\pi }{4})+cose{c}^2(x+\frac{\pi }{4})]dx$

$=e^x[-\cot (x+\frac{\pi }{4})]+c$ [Using $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+c$]

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