If $\int_0^b \frac{d x}{1+x^2}=\int_b^{\infty} \frac{d x}{1+x^2}$, then $b$ is equal to
If $\int_0^b \frac{d x}{1+x^2}=\int_b^{\infty} \frac{d x}{1+x^2}$, then $b$ is equal to
- $\tan ^{-1}\left(\frac{1}{3}\right)$
- $\frac{\sqrt{3}}{2}$
- $\sqrt{2}$
- $1$
Solution
We have,
$
\begin{aligned}
& \int_0^b \frac{d x}{1+x^2}=\int_b^{\infty} \frac{d x}{1+x^2} \\
\Rightarrow \quad & {\left[\tan ^{-1} x\right]_0^b=\left[\tan ^{-1} x\right]_b^{\infty} } \\
\Rightarrow & \tan ^{-1}(b)-\tan ^{-1}(0)=\tan ^{-1}(\infty)-\tan ^{-1}(b) \\
\Rightarrow \quad & 2 \tan ^{-1}(b)-0=\frac{\pi}{2} \\
\Rightarrow \quad & \tan ^{-1}(b)=\frac{\pi}{4} \Rightarrow b=\tan \left(\frac{\pi}{4}\right) \\
\therefore & b=1
\end{aligned}
$
Asked in: AP EAMCET 2013
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