Evaluate the following integral: `int_()^(pi//4)|tanx|dx`

Explanation

`I=int_{0}^{pi / y} tan ^{2} x(sec ^{2} x-1) d x`

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I=int_{0}^{pi / 4} {ten}^{2} x cdot sec ^{2} x d x-int_{0}^{pi / 4} tan ^{2} x d x`

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Let tan x=1 Rightarrow sec x {an}=dt`

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I=int_{0}^{1} t^{2} d t-int_{0}^{pi / 4}(sec ^{2} x-1) d x`

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.I=frac{[+3}{3}]_{0}^{1}-[tan x]_{0}^{pi / 4}`

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I=frac{1}{3}-1+frac{pi}{4}`

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I=frac{pi}{4}-frac{2}{3}`

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