Question

If $\int e^{\sin x}\left[\frac{x{\cos }^{3}x-\sin x}{{\cos }^{2}x}\right]dx=e^{\sin x}f\left(x\right)+c$ where c is constant of integration, then $f\left(x\right)=$

• A. $\sec x-x$
• B. $x-\sec x$
• C. $\tan x-x$
• D. $x-\tan x$

Answer 1

The correct option is B $x-\sec x$

$\int e^{\sin x}\left(\frac{x{\cos }^{3}x-\sin x}{{\cos }^{2}x}\right)dx=\int e^{\sin x}(x\cos x-\tan x\sec x)dx=\int x{e}^{\sin x}\cos xdx-\int e^{\sin x}\tan x\sec xdx$
Let
$I_1=\int x{e}^{\sin x}\cos xdx{I}_2=\int e^{\sin x}\tan x\sec xdx$
So using by parts,
$\Rightarrow I_1=x{e}^{\sin x}-\int e^{\sin x}.1dx$
And,
$\Rightarrow I_2=e^{\sin x}\sec x-\int \sec x\cdot e^{\sin x}\cos xdx\Rightarrow I_2=e^{\sin x}\sec x-\int e^{\sin x}dx$
Therefore,
$I_1-I_2=e^{\sin x}(x-\sec x)+c$

Hence $f\left(x\right)=x-\sec x$