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Evaluate: int0^(pi)(x)/(1+cos^2x)dx...

Evaluate: `int_0^(pi)(x)/(1+cos^2x)dx`

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`I=int_{0}^{pi} frac{x d x}{1+cos ^{2} x}`
` I=int_{0}^{pi} frac{pi-x}{1+cos ^{2}(pi-x)} d x(because int_{a}^{b} f(x) d x=int_{a}^{b} f(a+b-x) d x) `
` I=int_{0}^{pi} frac{pi d x}{1+cos ^{2} x}-I`
` therefore I=frac{pi}{2} int_{0}^{pi} frac{{dx}}{1+cos ^{2} {x}}`
` I=frac{pi}{2} cdot 2 cdot int_{0}^{pi / 2} frac{d x}{1+cos ^{2} x}( int_{0}^{2 a} f(x) d x=2 int_{0}^{a} f(x) d x. if .f(2 a-x)=f(x)) `
` I=pi int_{0}^{pi / 2} frac{d x}{1+cos ^{2} x}`
` I=pi int_{0}^{pi / 2} frac{sec ^{2} x}{sec ^{2} x+1} d x`
` =pi int_{0}^{pi / 2} frac{sec ^{2} x}{2+tan ^{2} x} d x`
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