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

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

Answers (1)

best_answer

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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