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  • Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 10.

Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 10.

Answers (1)

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

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: \int_{0}^{a}\sqrt{a^2-x^2}dx

Solution: \int_{0}^{a}\sqrt{a^2-x^2}dx

Put x=a\sin \theta

dx=a\cos \theta \; d\theta

When x=0  then \theta=0 and

 when x=a  then \theta=\frac{\pi}{2}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sqrt{a^{2}-a^{2} \sin ^{2} \theta} a \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{a^{2}\left(1-\sin ^{2} \theta\right)} a \cos \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1-\sin ^{2} x=\cos ^{2} x\right] \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} a \sqrt{\cos ^{2} \theta} a \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} a \cos \theta a \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} a^{2} \cos ^{2} \theta d \theta\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos ^{2} \theta=\frac{1+\cos 2 \theta}{2}\right] \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} a^{2}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \frac{a^{2}}{2}(1+\cos 2 \theta) d \theta \\ &=\frac{a^{2}}{2} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta \end{aligned}

\begin{aligned} &=\frac{a^{2}}{2}\left(\int_{0}^{\frac{\pi}{2}} 1 d \theta+\int_{0}^{\frac{\pi}{2}} \cos 2 \theta d \theta\right) \\ &=\frac{a^{2}}{2}[\theta]_{0}^{\frac{\pi}{2}}+\frac{a^{2}}{2}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{a^{2}}{2}\left[\frac{\pi}{2}-0\right]+\frac{a^{2}}{4}\left[\sin \frac{2 \pi}{2}-\sin 0\right] \end{aligned}

\begin{aligned} &=\frac{a^{2}}{2} \frac{\pi}{2}+\frac{a^{2}}{4}[\sin \pi-\sin 0] \\ &=\frac{a^{2} \pi}{4}+\frac{a^{2}}{4}[0-0] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\sin \pi=\sin 0=0] \end{aligned}

=\frac{\pi a^2}{4}

Posted by

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