Question

If $\int e^{{\sin }^{2}x}\sin x(\cos x+{\cos }^{3}x)dx=f\left(x\right)+C,$ where $C$ is a constant of integration, then $f\left(x\right)$ equals

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

Answer 1

The correct option is A $\frac{1}{2}{e}^{{\sin }^{2}x}(3-{\sin }^{2}x)$

$I=\int e^{{\sin }^{2}x}\sin x(\cos x+{\cos }^{3}x)dx$
$I=\int e^{{\sin }^{2}x}\sin x\cos xdx+\int e^{{\sin }^{2}x}\sin x{\cos }^{3}xdx$
Put ${\sin }^{2}x=t\Rightarrow 2\sin x\cos xdx=dt$
$I=\frac{1}{2}\int e^{t}dt+\frac{1}{2}\int e^t(1-t)dt$
$=\int e^{t}dt-\frac{1}{2}\int t{e}^{t}dt$
$=e^t-\frac{1}{2}[t{e}^t-e^t]+C$
$=\frac{3}{2}{e}^t-\frac{1}{2}t{e}^t+C$
$=\frac{1}{2}(3-{\sin }^{2}x)e^{{\sin }^{2}x}+C$