Evaluate the following: ^ 2 _0 dx 1+m^2 ^2x dx

Question

Evaluate the following:
$\int\limits^{\frac{\pi}{2}}_0\frac{\tan\text{x dx}}{1+\text{m}^2\tan^2\text{x}}\text{dx}$

✓

Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan\text{x dx}}{1+\text{m}^2\tan^2\text{x}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\frac{\sin\text{x}}{\cos\text{x}}}{1+\text{m}^2\cdot\frac{\sin^2\text{x}}{\cos^2\text{x}}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\text{x}\cos\text{x dx}}{1-\sin^2\text{x}+\text{m}^2\sin^2\text{x}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\text{x}\cos\text{x}}{1+(\text{m}^2-1)\sin^2\text{x}}\text{dx}$
Put $\sin^2\text{x}=\text{t}$ $\Rightarrow\ 2\sin\text{x}\cos\text{x}\text{ dx}=\text{dt}$
When $\text{x}\rightarrow0,$ then $\text{t}\rightarrow0$ and $\text{x}\rightarrow\frac{\pi}{2},$ then $\text{t}\rightarrow1$
$\therefore\ \text{I}=\frac{1}{2}\int\limits^1_0\frac{\text{dt}}{1+(\text{m}^2-1)\text{t}}$
$=\frac{1}{2}\Big[\frac{1}{\text{m}^2-1}\log(1+(\text{m}^2-1)\text{t})\Big]^1_0$
$=\frac{1}{2(\text{m}^2-1)}\big[\log(1+(\text{m}^2-1))-\log1\big]$
$=\frac{\log\text{m}^2}{2(\text{m}^2-1)}=\frac{2\log|\text{m}|}{2(\text{m}^2-1)}=\frac{\log|\text{m}|}{\text{m}^2-1}$

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