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  • Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 46.

Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 46.

Answers (1)

Answer :  \frac{8}{15}

Hint: Use indefinite formula and the given limits to solve this integral
Given:

\int_{0}^{\frac{\pi}{2}}\cos ^5x \; dx

Solution:

\begin{aligned} \int_{0}^{\pi / 2} \cos ^{5} x d x &=\int_{0}^{\pi / 2} \cos ^{4} x \cdot \cos x d x \\ &=\int_{0}^{\pi / 2}\left(\cos ^{2} x\right)^{2} \cos x d x \\ &=\int_{0}^{1 / 2}\left(1-\sin ^{2} x\right)^{2} \cos x d x \quad \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore 1=\sin ^{2} \theta+\cos ^{2} \theta\right] \end{aligned}

Put  \sin x=t\Rightarrow \cos x \; dx=dt

when x=0  then t=0

whenx=\frac{\pi}{2}  then t=1

\begin{aligned} &\therefore \int_{0}^{\pi / 2} \cos ^{5} x d x=\int_{0}^{1}\left(1-t^{2}\right)^{2} d t \\ &\qquad \begin{aligned} & \int_{0}^{1}\left(1+t^{4}-2 t^{2}\right) d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\ =& \int_{0}^{1} t^{0} d t+\int_{0}^{1} t^{4} d t-2 \int_{0}^{1} t^{2} d t \end{aligned} \end{aligned}

\begin{aligned} &=\left[\frac{t^{1+1}}{0+1}\right]_{0}^{1}+\left[\frac{t^{4+1}}{4+1}\right]_{0}^{1}-2\left[\frac{t^{2+1}}{2+1}\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \because \int x^{n} d x=\frac{x^{n+1}}{x+1}+c \\ &=[t]_{0}^{1}+\frac{1}{5}\left[t^{5}\right]_{0}^{1}-\frac{2}{3}\left[t^{3}\right]_{0}^{1} \\ &=[1-0]+\frac{1}{5}\left[1^{5}-0^{5}\right]-\frac{2}{3}\left[1^{3}-0^{3}\right] \\ &=1+\frac{1}{5}-\frac{2}{3} \\ &=\frac{15+3-10}{15} \end{aligned}

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