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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\tan^{2}\text{x}}{1+\text{e}^{\text{x}}}\text{ dx}$
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Answer

We now $\int_\limits{a}^{b}\text{f}\text{(x)}\text{dx}=\int_\limits{a}^{b} \text{f}(\text{a}+\text{b}-\text{x}) \text{dx}$
Hence,
$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{(-x)}}{1-\text{e}^\text{-x}}\text{dx}$
$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1-\text{e}^\text{-x}}\text{dx}$
If,
$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}$
Then
$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{-x}}\text{dx}$
So,
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\text{e}^x\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^2}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$.
$\text{I}=\frac{1}{2}\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$
We know
If f(x)is even
$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=2\int_\limits{0}^{a}\text{f}\text{(x)}\text{dx}$
If f(x)is odd
$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=0$
Here
$\text{f}\text{(x)}=\tan^2\text{x}$
f(x)is even,hence
$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$
$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\sec^2\text{x}-1\text{dx}$.
$\text{I}=\big\{\tan\text{x}-\text{x}\big\}\frac{\frac{\pi}{4}}{0 }$
$\text{I}=1-\frac{\pi}{4}$

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