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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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