Evaluate the following integrals: ^ 2 _ 0 1+m^2 ^2x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$

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Answer

Let $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\frac{\sin\text{x}}{\cos\text{x}}}{1+\text{m}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}}\text{ dx}=\int^\limits{\frac{\pi}{2}}_{0}\frac{{\sin\text{x}}{\cos\text{x}}}{\cos^2\text{x}+\text{m}^2{\sin^2\text{x}}}\text{ dx}$
Put $\cos^2\text{x}+\text{m}^2\sin^2\text{x}=\text{z}$
$\therefore\ 2\cos\text{x}(-\sin\text{x})\text{dx}+\text{m}^2\times2\sin\text{x}\cos\text{x dx}=\text{ dz}$
$\Rightarrow2(\text{m}^2-1)\sin\text{x}\cos\text{x dx}=\text{dz}$
$\Rightarrow\sin\text{x}\cos\text{x dx}=\frac{\text{dz}}{2(\text{m}^2-1)}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=1+\text{m}^2\times0=1\big)$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\text{m} ^2$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=0+\text{m}^2\times0=\text{m}^2\big)$
$\therefore\ \text{I}=-\frac{1}{2(\text{m}^2-1)}\int\limits^{\text{m}^2}_1\frac{\text{dz}}{\text{z}}$
$=\frac{1}{2(\text{m}^2-1)}\big[\log\text{z}\big]^{\text{m}^2}_1$
$=\frac{1}{2(\text{m}^2-1)}\big(\log\text{m}^2-\log1\big)$
$=\frac{1}{2(\text{m}^2-1)}\big(2\log|\text{m}|-0\big)$
$=\frac{\log|\text{m}|}{\text{m}^2-1}$

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