#### Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 11.

Answer$\frac{64}{231}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given:$\int_{0}^{\frac{\pi}{2}}\sqrt{\sin \phi}\cos^5 \phi d\phi$

Solution: $\int_{0}^{\frac{\pi}{2}}\sqrt{\sin \phi}\cos^5 \phi d\phi$

Put $\sin \phi=t$

$\cos \phi d\phi =dt$

$d\phi =\frac{dt}{cos\phi}$

When $\phi =0$  then $t=0$  and when $\phi =\frac{\pi}{2}$  then $t=1$

\begin{aligned} &=\int_{0}^{1} \sqrt{t} \cos ^{5} \phi \frac{d t}{\cos \phi} \\ &=\int_{0}^{1} \sqrt{t} \cos ^{4} \phi d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(\cos ^{2} \phi\right)^{2} d t \end{aligned}

\begin{aligned} &=\int_{0}^{1} t^{\frac{1}{2}}\left(1-\sin ^{2} \phi\right)^{2} d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(1-t^{2}\right)^{2} d t \\ &=\int_{0}^{1} t^{\frac{1}{2}}\left(1+t^{4}-2 t^{2}\right) d t \end{aligned}

\begin{aligned} &=\int_{0}^{1}\left(t^{\frac{1}{2}}+t^{4+\frac{1}{2}}-2 t^{2+\frac{1}{2}}\right) d t \\ &=\int_{0}^{1} t^{\frac{1}{2}} d t+\int_{0}^{1} t^{\frac{9}{2}} d t-2 \int_{0}^{1} t^{\frac{5}{2}} d t \end{aligned}

\begin{aligned} &=\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{9}{2}+1}}{\frac{9}{2}+1}-\frac{2 t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{0}^{1} \\ &=\left[\frac{t^{\frac{3}{2}}}{3}\right]_{0}^{1}+\left[\frac{t^{\frac{11}{2}}}{\frac{11}{2}}\right]_{0}^{1}-2\left[\frac{t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{11}\left[t^{\frac{11}{2}}\right]_{0}^{1}-2 \times \frac{2}{7}\left[t^{\frac{7}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]+\frac{2}{11}\left[1^{\frac{11}{2}}-0^{\frac{11}{2}}\right]-\frac{4}{7}\left[1^{\frac{7}{2}}-0^{\frac{7}{2}}\right] \\ &=\frac{2}{3}[1-0]+\frac{2}{11}[1-0]-\frac{4}{7}[1-0] \end{aligned}

$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$

$=\frac{154+42-132}{231}$

$=\frac{196-132}{231}$

$=\frac{64}{231}$