Question

Solve $\underset{0}{\overset{\pi /2}{\int }}{\sin }^{4}x{\cos }^{3}xdx$

• A. $\frac{6}{35}$
• B. $\frac{2}{21}$
• C. $\frac{2}{15}$
• D. $\frac{2}{35}$

Answer 1

Answer: (D)

$\frac{2}{35}$

${\int }_0^{\pi /2}{\sin }^{4}x{\cos }^{3}xdx$

${\int }_0^{\pi /2}{\sin }^{4}x(1-{\sin }^{2}x)\cos xdx$ $\because {\cos }^{2}x=1-{\sin }^{2}x$

${\int }_0^{\pi /2}({\sin }^{4}x-{\sin }^{6}x)\cos xdx$

substitute $\sin x=t$ hence, $dt=\cos xdx$

and at $x=0;t=\sin \left(0\right)=0;$ at $x=\pi /2;t=\sin \left(\pi /2\right)=1$

${\int }_0^1(t^4-t^6)dt$

$[\frac{t^5}{5}-\frac{t^7}{7}{]}_0^1$

$[\frac{1}{5}-\frac{1}{7}]=\frac{2}{35}$