Question

$\int e^{\sin x}(x\cos x-\sec x\tan x)dx=$ _______ $+C;0<x<\frac{\pi }{2}$.

• A. $e^{\sin x}(x-\sec x)$
• B. $e^{\sin x}(\sec x-x)$
• C. $e^{\sin x}x\cos x$
• D. $e^{\sin x}(x+\sec x)$

Answer 1

Answer: (A)

$e^{\sin x}(x-\sec x)$

Direct integration is long and tedious here. Hence, we use completion of differential terms as the integrand looks like it has the form $d\left(uv\right)=udv+vdu$

$\therefore \int e^{\sin x}(x\cos x-\sec x\tan x)dx=\int e^{\sin x}(x\cos x-1+1-\sec x\tan x)dx$

$=\int e^{\sin x}(x\cos x-\sec x\cos x+1-\sec x\tan x)dx$

$=\int e^{\sin x}\left(\cos x\right(x-\sec x)dx+d(x-\sec x\left)\right)$ (as $d(x-\sec x)=1-\sec x\tan x$

$=\int (x-\sec x)d\left(e^{\sin x}\right)+e^{\sin x}d(x-\sec x)$(as $d\left(e^{\sin x}\right)=e^{\sin x}\cos x$)

$=\int d\left[e^{\sin x}\right(x-\sec x\left)\right]$ (as $d\left(uv\right)=vdu+udv$)

$=e^{\sin x}(x-\sec x)+C$

This is the required answer.