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

Answers (1)

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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