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  • Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 34 maths textbook solution

Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 34 maths textbook solution

Answers (1)

Answer: a=2

Hint: you must know the rule of integration

Given: -\int_{0}^{a} \frac{1}{4+x^{2}} d x=\frac{\pi}{8}, \text { find } a

Solution:  

\begin{aligned} &\int_{0}^{a} \frac{1}{4+x^{2}} d x=\frac{\pi}{8} \\\\ &\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a} \end{aligned}

\begin{aligned} &\therefore \int_{0}^{a} \frac{1}{2^{2}+x^{2}} d x=\frac{\pi}{8} \\\\ &\Rightarrow \frac{1}{2} \tan ^{-1} \frac{a}{2}-0=\frac{\pi}{8} \end{aligned}

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

\Rightarrow a=2

 

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