Question

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

Answer 1

\[\text{Let I }= \int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

\[= \int_0^\frac{\pi}{2} \frac{\frac{\sin x}{\cos x}}{1 + m^2 \frac{\sin^2 x}{\cos^2 x}}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\sin x\cos x}{\cos^2 x + m^2 \sin^2 x}dx\]

Put

\[\cos^2 x + m^2 \sin^2 x = z\]

\[\therefore 2\cos x\left( - \sin x \right)dx + m^2 \times 2\sin x\cos\ x\ dx = dz\]
\[ \Rightarrow 2\left( m^2 - 1 \right)\sin x\cos\ x\ dx = dz\]
\[ \Rightarrow \sin x\cos\ x\ dx = \frac{dz}{2\left( m^2 - 1 \right)}\]

When

\[x \to 0, z \to 1\]

\[\left( z = \cos^2 x + m^2 \sin^2 x = 1 + m^2 \times 0 = 1 \right)\]

When

\[x \to \frac{\pi}{2}, z \to m^2\]

\[\left( z = \cos^2 x + m^2 \sin^2 x = 0 + m^2 \times 1 = m^2 \right)\]

\[\therefore I = \frac{1}{2\left( m^2 - 1 \right)} \int_1^{m^2} \frac{dz}{z}\]
\[ = \left.\frac{1}{2\left( m^2 - 1 \right)} \log z\right|_1^{m^2} \]
\[ = \frac{1}{2\left( m^2 - 1 \right)}\left( \log m^2 - \log1 \right)\]
\[ = \frac{1}{2\left( m^2 - 1 \right)}\left( 2\log\left| m \right| - 0 \right)\]
\[ = \frac{\log\left| m \right|}{m^2 - 1}\]