Question

Evaluate $\int \left(\tan \right(e^x)+x{e}^x.{\sec }^2(e^x\left)\right)dx$

• A. $x-\tan \left(e^x\right)+c$
• B. $x\tan \left(e^x\right)+c$
• C. $x\tan \left(e^{-x}\right)+c$
• D. $x-\tan \left(e^{-x}\right)+c$

Answer 1

Answer: (B)

$x\tan \left(e^x\right)+c$

$I=\int \left(\tan \right(e^x)+x{e}^x.{\sec }^2(e^x\left)\right)dx$

$=\int \tan \left(e^x\right)dx+\int x{e}^x.{\sec }^2\left(e^x\right)dx=I_1+I_2$

$I_1=\int 1.\tan \left(e^x\right)dx$

We have,

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$I_1=x\tan \left(e^x\right)-\int \frac{d\left\{\tan \right(e^x\left)\right\}}{dx}x.dx+c$, Using integral by parts.

$=x\tan \left(e^x\right)-\int x{e}^x.{\sec }^2\left(e^x\right)dx+c=x\tan \left(e^x\right)-I_2+c$

$\Rightarrow I=I_1+I_2=x\tan \left(e^x\right)+c$