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

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}

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