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प्रश्न
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उत्तर
\[\text{Let I }= \int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\sin x\cos x}{\cos^2 x + m^2 \sin^2 x}dx\]
Put
\[\therefore 2\cos x\left( - \sin x \right)dx + m^2 \times 2\sin x\cos\ x\ dx = dz\]
\[ \Rightarrow 2\left( m^2 - 1 \right)\sin x\cos\ x\ dx = dz\]
\[ \Rightarrow \sin x\cos\ x\ dx = \frac{dz}{2\left( m^2 - 1 \right)}\]
When
When
\[\therefore I = \frac{1}{2\left( m^2 - 1 \right)} \int_1^{m^2} \frac{dz}{z}\]
\[ = \left.\frac{1}{2\left( m^2 - 1 \right)} \log z\right|_1^{m^2} \]
\[ = \frac{1}{2\left( m^2 - 1 \right)}\left( \log m^2 - \log1 \right)\]
\[ = \frac{1}{2\left( m^2 - 1 \right)}\left( 2\log\left| m \right| - 0 \right)\]
\[ = \frac{\log\left| m \right|}{m^2 - 1}\]
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