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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 11

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Answer: \frac{\pi }{2}

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

Given: \int_{0}^{4} \frac{1}{\sqrt{16-x^{2}}} d x

Solution:  \int_{0}^{4} \frac{1}{\sqrt{16-x^{2}}} d x

=\int_{0}^{4} \frac{1}{\sqrt{(4)^{2}-x^{2}}} d x

\begin{aligned} &\text { Put } x=4 \sin \theta \quad \theta=\sin ^{-1} \frac{x}{4} \\ &\mathrm{~d} \mathrm{x}=4 \cos \theta\; d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{4} \frac{4 \cos \theta}{\sqrt{16-16 \sin ^{2} \theta}} d \theta \\\\ &=\int_{0}^{4} \frac{4 \cos \theta}{\sqrt{16\left(1-\sin ^{2} \theta\right)}} d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{4} \frac{4 \cos \theta}{4 \sqrt{\cos ^{2} \theta}} d \theta \\\\ &=\int_{0}^{4} \frac{4 \cos \theta}{4 \cos \theta} d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{4} 1 d \theta \\\\ &=[\theta]_{0}^{4}=\left[\operatorname{Sin}^{-1} \frac{x}{4}\right]_{0}^{4} \end{aligned}

\begin{aligned} &=\left[\sin ^{-1} \frac{4}{4}-\sin ^{-1} \frac{0}{4}\right] \\\\ &=\sin ^{-1} 1 \\\\ &=\frac{\pi}{2} \end{aligned}


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