Introduction to definite integration

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Question

$\int_{0}^{π / 4} (\tan^{4} x + \tan^{2} x) d x$

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Solution

$\begin{array}{l} G.I = \int_{0}^{π / 4} (\tan^{2} x + 1) \tan^{2} x d x \\ = \int_{0}^{π / 4} \tan^{2} x \cdot \sec^{2} x d x \\ = {[\frac{\tan^{3} x}{3}]}_{0}^{π / 4} = \frac{1}{3} = 0.3330 \end{array}$

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