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प्रश्न
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उत्तर
\[\int_0^\frac{\pi}{4} \tan^2 x\ d x\]
\[ = \int_0^\frac{\pi}{4} \left( se c^2 x - 1 \right) d x\]
\[ = \left[ \tan x - x \right]_0^\frac{\pi}{4} \]
\[ = 1 - \frac{\pi}{4} - 0\]
\[ = 1 - \frac{\pi}{4}\]
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