Question
If $\int\limits^\alpha_0\frac{1}{1+4\text{x}^2}\text{ dx}=\frac{\pi}{8},$ then a equals:
-
$\frac{\pi}{2}$
-
$\frac{1}{2}$
-
$\frac{\pi}{4}$
-
$1$
$\frac{\pi}{2}$
$\frac{1}{2}$
$\frac{\pi}{4}$
$1$
Solution:
$\int\limits^\alpha_0\frac{1}{1+4\text{x}^2}\text{ dx}=\frac{\pi}{8}$
$\Rightarrow\int\limits^\alpha_0\frac{1}{1+(2\text{x)}^2}\text{ dx}=\frac{\pi}{8}$
$\Rightarrow \frac{1}{2}\big[\tan^-2\text{x}\big]^\alpha_0=\frac{\pi}{8}$
$\Rightarrow \frac{1}{2}\tan^{-1}2\alpha=\frac{\pi}{8}$
$\Rightarrow 2\alpha=\tan\frac{\pi}{4}$
$\Rightarrow2\alpha=1$
$\therefore\alpha=\frac{1}{2}$
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