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  • Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 43

Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 43

Answers (1)

best_answer

Answer:

1

Hint:

To solve this equation.

Given:

\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+\cos 2 x} d x

Explanation:

\begin{aligned} &\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+\cos 2 x} d x \\\\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 \cos ^{2} x} d x \\\\ &1+\cos 2 x=2 \cos ^{2} x \end{aligned}

\begin{aligned} &=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\cos ^{2} x} d x \\\\ &=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x \\\\ &\frac{1}{\cos x}=\sec x \end{aligned}

\begin{aligned} &=\frac{1}{2}[\tan x]_{\frac{\pi}{4}}^{\frac{\pi}{4}} \\\\ &=\frac{1}{2}\left[\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)\right] \\\\ &=\frac{1}{2}[1-(-1)] \end{aligned}

\begin{aligned} &=\frac{1}{2}[1+1] \\\\ &=\frac{2}{2} \\\\ &=1 \end{aligned}

 

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