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  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 7 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 7 maths textbook solution.

Answers (1)

Answer:-  \frac{\pi}{4}

Hints:-  You must know the integration rules of trigonometric functions and its limits

Given:-  \int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} d x

Solution : \int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} d x

x=a \sin \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[x=0, a \sin \theta=0, \theta=0]

d x=a \cos \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[x=a, 0 \sin \theta=a, \sin \theta=1, \theta=\pi / 2]

I=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a \sin \theta+\sqrt{a^{2}-a^{2} \sin ^{2} \theta}} d \theta

\begin{aligned} &=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a \sin \theta+a \sqrt{1-\sin ^{2} \theta}} d \theta \\ &=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a \sin \theta+a \cos \theta} d \theta \\ &=\int_{0}^{\pi / 2} \frac{a \cos \theta}{a(\sin \theta+\cos \theta)} d \theta \end{aligned}

=\int_{0}^{\pi / 2} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta                                                           .....(1)

I=\int_{0}^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\sin \left(\frac{\pi}{2}-\theta\right)+\cos \left(\frac{\pi}{2}-\theta\right)} d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]

I=\int_{0}^{\pi / 2} \frac{\sin \theta}{\cos \theta+\sin \theta} d \theta                                                       .....(2)

Adding both (1) and (2)

\begin{aligned} &I+I=\int_{0}^{\pi / 2} \frac{\sin \theta}{\cos \theta+\sin \theta} d \theta+\int_{0}^{\pi / 2} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta \\ &2 I=\int_{0}^{\pi / 2} \frac{\sin \theta+\cos \theta}{\sin \theta+\cos \theta} d \theta \\ &2 I=\int_{0}^{\pi / 2} 1 . d \theta \end{aligned}

\begin{aligned} &2 I=[0]_{0}^{\pi / 2} \\ &2 I=\left[\frac{\pi}{2}-0\right] \\ &I=\frac{\pi}{4} \end{aligned}

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