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  • Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 17 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 17 textbook solution.

Answers (1)

Answer:- \frac{\pi^{2}}{4}

Hints:-  You must know about the integration rules of trigonometry functions.

Given:-  \int_{0}^{\pi} x \cos ^{2} x d x

Solution :  \int_{0}^{\pi} x \cos ^{2} x d x=I

\begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \cos ^{2}(\pi-x) d x \\ &I=\int_{0}^{\pi}(\pi-x) \cos ^{2} x \cdot d x \\ &I=\int_{0}^{\pi} \pi \cos ^{2} x \cdot d x-\int_{0}^{\pi} x \cos ^{2} x \cdot d x \end{aligned}

\begin{aligned} &I=\pi \int_{0}^{\pi} \cos ^{2} x \cdot d x-I \\ &2 I=\pi \int_{0}^{\pi} \frac{\cos 2 x+1}{2} \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{\cos 2 x}{2} \cdot d x+\int_{0}^{\pi} \frac{1}{2} \cdot d x \end{aligned}

\begin{gathered} =\pi\left[\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\pi}+\frac{1}{2}[x]_{0}^{\pi}\right] \\ 2 I=\frac{\pi}{4}[\sin 2 \pi-\sin 0]+\frac{\pi}{2}[\pi] \end{gathered}

\begin{aligned} &2 I=\frac{\pi^{2}}{4} \\ &I=\frac{\pi^{2}}{4} \end{aligned}

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