Question

$\int {\cos }^{3}x.e^{\log \sin x}dx$ is equal to

• A. $-\frac{{\sin }^{4}x}{4}+c$
• B. $-\frac{{\cos }^{4}x}{4}+c$
• C. $\frac{e^{\sin x}}{4}+c$
• D. None of these

Answer 1

Answer: (B)

$-\frac{{\cos }^{4}x}{4}+c$

Let $I=\int {\cos }^{3}x.e^{\log \sin x}dx=\int {\cos }^3\sin xdx$
Put $\cos x=t\Rightarrow -\sin xdx=dt$
$\therefore I=-{\int }^{}{t}^{3}dt=0\frac{t^4}{4}+c=-\frac{{\cos }^{4}x}{4}+c$