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

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

Answers (1)

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}

Posted by

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