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प्रश्न
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उत्तर
Let
\[= \int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\cos^6 x \left( \tan^3 x + 1 \right)^2}dx\]
\[ = \int_0^\frac{\pi}{4} \frac{\tan^2 x \sec^2 x}{\left( \tan^3 x + 1 \right)^2}dx\]
Put
\[\therefore 3 \tan^2 x \sec^2 xdx = dz\]
\[ \Rightarrow \tan^2 x \sec^2 xdx = \frac{dz}{3}\]
When
When
\[\therefore I = \frac{1}{3} \int_1^2 \frac{dz}{z^2}\]
\[ = \left.\frac{1}{3} \times - \frac{1}{z}\right|_1^2 \]
\[ = - \frac{1}{3}\left( \frac{1}{2} - 1 \right)\]
\[ = - \frac{1}{3} \times \left( - \frac{1}{2} \right)\]
\[ = \frac{1}{6}\]
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