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  • Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 15 textbook solution.

Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 15 textbook solution.

Answers (2)

Answer : \frac{\pi}{12}

Given : \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x

Hint : Use the formula of \int_{a}^{b} f(x) d x

Solution : I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x

I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x----(1)\left(\tan x=\frac{\sin x}{\cos x}\right)

We know that \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x-------(2) \end{aligned}

Add (1) and (2)

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 . d x \\ &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \end{aligned}

\begin{aligned} &2 I=\frac{2 \pi-\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}

 

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Answer : \frac{\pi}{12}

Given : \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x

Hint : Use the formula of \int_{a}^{b} f(x) d x

Solution : I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} d x

I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x----(1)\left(\tan x=\frac{\sin x}{\cos x}\right)

We know that \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x

\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x \\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x-------(2) \end{aligned}

Add (1) and (2)

\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\ &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 . d x \\ &2 I=(x)_{\frac{\pi}{6}}^{\frac{\pi}{3}} \end{aligned}

\begin{aligned} &2 I=\frac{2 \pi-\pi}{6} \\ &I=\frac{\pi}{12} \end{aligned}

 

Posted by

infoexpert23

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