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  • Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 17 maths textbook solution

Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 17 maths textbook solution

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Answer: 0

Hint: You must know the integration rules of trigonometric function with its limits


Given: \int_{0}^{\pi} \cos ^{5} x \; d x

Solution:  Using property \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x

\mathrm{I}=\int_{0}^{\pi} \cos ^{5} x \; d x=\int_{0}^{\pi} \cos x \cos ^{4} x \; d x

    =\int_{0}^{\pi} \cos x\left(1-\sin ^{2} x\right)^{2} d x

\begin{aligned} &\mathrm{Put} \sin \mathrm{x}=\mathrm{t} \\\\ &\cos x \; d x=d t \end{aligned}

\begin{aligned} &\mathrm{I}=\int_{0}^{\pi}\left(1-t^{2}\right)^{2} d x \\\\ &=\int_{0}^{\pi}\left(1+t^{4}-2 t^{2}\right) d x \end{aligned}

=\left[t+\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]_{0}^{\pi}

\begin{aligned} &=\left[\sin x+\frac{\sin ^{5} x}{5}-\frac{2 \sin ^{3} x}{3}\right]_{0}^{\pi} \quad\quad\quad\quad[\sin \pi=0, \sin 0=0] \\\\ &I=0 \end{aligned}

 

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