Question
Fill in the blanks:
If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then a = ________.
If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then a = ________.
✓
Answer
If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then $\text{a}=\frac{1}{2}$
Let $\text{I}=\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8}$
Now, $\int\limits^\text{a}_0\frac{1}{4\Big(\frac{1}{4}+\text{x}^2\Big)}\text{dx}=\frac{2}{4}\big[\tan^{-1}2\text{x}\big]^\text{a}_0$
$=\frac{1}{2}\tan^{-1}2\text{a}-0=\frac{\pi}{8}$
$\frac{1}{2}\tan^{-1}2\text{a}=\frac{\pi}{8}$
$\Rightarrow\ \tan^{-1}2\text{a}=\frac{\pi}{4}$
$\Rightarrow\ 2\text{a}=1$
$\therefore\ \text{a}=\frac{1}{2}$
Let $\text{I}=\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8}$
Now, $\int\limits^\text{a}_0\frac{1}{4\Big(\frac{1}{4}+\text{x}^2\Big)}\text{dx}=\frac{2}{4}\big[\tan^{-1}2\text{x}\big]^\text{a}_0$
$=\frac{1}{2}\tan^{-1}2\text{a}-0=\frac{\pi}{8}$
$\frac{1}{2}\tan^{-1}2\text{a}=\frac{\pi}{8}$
$\Rightarrow\ \tan^{-1}2\text{a}=\frac{\pi}{4}$
$\Rightarrow\ 2\text{a}=1$
$\therefore\ \text{a}=\frac{1}{2}$
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