Question-217881

Question Number 217881 by OmoloyeMichael last updated on 23/Mar/25

Answered by vnm last updated on 23/Mar/25

\int_{0}^{π / 2} x (\frac{\sin x}{1 + \cos^{2} x} + \frac{\cos x}{1 + \sin^{2} x}) d x = I

f (x) = \frac{\sin x}{1 + \cos^{2} x} + \frac{\cos x}{1 + \sin^{2} x}

F (x) = \int f (x) d x = - \int \frac{d (\cos x)}{1 + \cos^{2} x} + \int \frac{d (\sin x)}{1 + \sin^{2} x} =

- \tan^{- 1} (\cos x) + \tan^{- 1} (\sin x)

I = x F (x) ∣_{0}^{π / 2} - \int_{0}^{π / 2} {(x)}^{'} F (x) d x = \frac{π^{2}}{8} - \int_{0}^{π / 2} F (x) d x

F (x) \underset{0 ⩽ x ⩽ \frac{π}{2}}{=} - F (\frac{π}{2} - x) \Rightarrow \int_{0}^{π / 2} F (x) d x = 0

I = \frac{π^{2}}{8}

Answered by mr W last updated on 23/Mar/25

I = \int_{0}^{\frac{π}{2}} x (\frac{\sin x}{1 + \cos^{2} x} + \frac{\cos x}{1 + \sin^{2} x}) d x = I_{1} + I_{2}

I_{1} = \int_{0}^{\frac{π}{2}} x (\frac{\sin x}{1 + \cos^{2} x}) d x

I_{2} = \int_{0}^{\frac{π}{2}} x (\frac{\cos x}{1 + \sin^{2} x}) d x

= \int_{0}^{\frac{π}{2}} (x - \frac{π}{2} + \frac{π}{2}) (\frac{\cos (x - \frac{π}{2} + \frac{π}{2})}{1 + \sin^{2} (x - \frac{π}{2} + \frac{π}{2})}) d (x - \frac{π}{2})

= \int_{- \frac{π}{2}}^{0} (t + \frac{π}{2}) (\frac{\cos (t + \frac{π}{2})}{1 + \sin^{2} (t + \frac{π}{2})}) d t

= \int_{0}^{- \frac{π}{2}} (t + \frac{π}{2}) (\frac{\sin t}{1 + \cos^{2} t}) d t

= \int_{0}^{- \frac{π}{2}} t (\frac{\sin t}{1 + \cos^{2} t}) d t + \frac{π}{2} \int_{0}^{- \frac{π}{2}} (\frac{\sin t}{1 + \cos^{2} t}) d t

= - I_{1} + \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin t}{1 + \cos^{2} t} d t

= - I_{1} - \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{d (\cos t)}{1 + \cos^{2} t}

= - I_{1} - \frac{π}{2} {[\tan^{- 1} (\cos t)]}_{0}^{\frac{π}{2}}

= - I_{1} - \frac{π}{2} (0 - \frac{π}{4})

= - I_{1} + \frac{π^{2}}{8}

\Rightarrow I = I_{1} + I_{2} = \frac{π^{2}}{8}

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