Question

$\int e^{\tan x}\frac{1}{{\cos }^{4}x}dx$ is equal to

• A. $(\tan x+1{)}^2{e}^x+C$
• B. $({\tan }^{2}x-2\tan x+3)e^{\tan x}+C$
• C. $({\tan }^{2}x+2\tan x+2)e^{\tan x}+C$
• D. $({\tan }^{2}x-2\tan x+4)e^{\tan x}+C$

Answer 1

The correct option is B $({\tan }^{2}x-2\tan x+3)e^{\tan x}+C$

$\int e^{\tan x}\frac{1}{{\cos }^{4}x}dx=\int e^{\tan x}\frac{({\sin }^{2}x+{\cos }^{2}x)}{{\cos }^{4}x}dx=\int e^{\tan x}({\tan }^{2}x+1){\sec }^{2}xdx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\int e^t(t^2+1)dt=(t^2+1)e^t-\int \left(2t\right)e^{t}dt+c_1=(t^2+1)e^t-\left(2t\right)e^t+\int 2e^{t}dt+c_2=(t^2+1-2t+2)e^t+C=(t^2-2t+3)e^t+C=({\tan }^{2}x-2\tan x+3)e^{\tan x}+C$