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#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 1 maths textbook solution

Answer: $a=\frac{1}{2}$

Hint: You must know about the $\int_{a}^{b} \frac{1}{1+x^{2}} d x$

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

Solution:

$I=\int_{0}^{a} \frac{1}{1+4 x^{2}} d x$

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

$=\frac{1}{4} \int_{0}^{a} \frac{1}{x^{2}+\left(\frac{1}{2}\right)^{2}} d x$                    $\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]$

$=\frac{1}{4} \cdot \frac{1}{1 / 2}\left[\tan ^{-1} \frac{x}{1 / 2}\right]_{0}^{a}$

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

$=\frac{1}{2} \tan ^{-1}(2 a)$

According to given,

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

$\Rightarrow(2 a)=\tan \frac{\pi}{4} \Rightarrow 2 a=1 \quad \Rightarrow a=\frac{1}{2} \quad\left[\because \tan \frac{\pi}{4}=1\right]$