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  • Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 21 maths textbook solution

Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 21 maths textbook solution

Answers (1)

Answer:

\frac{1}{2}

Given:

\int_{0}^{a} \frac{1}{1+4 x^{2}} d x=\frac{\pi}{8}

Hint:

To solve this equation, we will do 4x in whole system.
 

Explanation:  

Let

\begin{aligned} &I=\int_{0}^{a} \frac{1}{1+4 x^{2}} d x \\\\ &I=\int_{0}^{a} \frac{1}{1+(2 x)^{2}} d x \end{aligned}

\begin{aligned} &=\left[\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1}\right)\right]_{0}^{a} \\\\ &=\frac{\pi}{8} \\\\ &\frac{1}{2}\left[\tan ^{-1}(2 a)-\tan ^{-1}(0)\right]=\frac{\pi}{8} \end{aligned}

\begin{aligned} &\frac{1}{2}\left[\tan ^{-1}(2 a)\right]=\frac{\pi}{8} \\\\ &\tan ^{-1}(2 a)=\frac{\pi}{4} \end{aligned}

\begin{aligned} &2 a=\tan \frac{\pi}{4} \\\\ &2 a=1 \\\\ &a=\frac{1}{2} \end{aligned}

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