Evaluate: $\int_{0}^{\frac{π}{2}} 2 \sin x \cos x \tan^{- 1} (\sin x) dx$

Solution

Consider the given integral

$I = \int_{0}^{\frac{π}{2}} 2 \sin x \cos x \tan^{- 1} (\sin x) dx Let t = \sin x \Rightarrow dt = \cos x dx When x = \frac{π}{2}, t = 1 When x = 0, t = 0 Now, \int 2 \sin x \cos x \tan^{- 1} (\sin x) dx = \int 2 t \tan^{- 1} t dt$

$= [\tan^{- 1} t] \int 2 t dt - \int [\frac{d}{dt} . (\tan^{- 1}) \int 2 t dt] dt = [\tan^{- 1} t] [2 . \frac{t^{2}}{2}] - \int (\frac{1}{1 + t^{2}} x 2 . \frac{t^{2}}{2}) dt = t^{2} \tan^{- 1} t - \int \frac{t^{2}}{1 + t^{2}} dt = t^{2} \tan^{- 1} t - \int [1 - \frac{1}{1 + t^{2}}] dt = t^{2} \tan^{- 1} t - t + \tan^{- 1} t$

$∴ I = \int_{0}^{\frac{π}{2}} 2 \sin x \cos x \tan^{- 1} (\sin x) dx = {[t^{2} \tan^{- 1} t - t + \tan^{- 1} t]}_{0}^{1} = [1^{2} \tan^{- 1} 1 - 1 + \tan^{- 1} 1] - [0^{2} \tan^{- 1} 0 - 0 + \tan^{- 1} 0] = [1 x \frac{π}{4} - 1 + \frac{π}{4}] - 0 = \frac{π}{4} - 1 + \frac{π}{4} = \frac{π}{2} - 1$

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Integrals

Evaluate: $\int_{0}^{\frac{π}{2}} 2 \sin x \cos x \tan^{- 1} (\sin x) dx$

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Evaluate: $\int_{0}^{\frac{π}{2}} 2 \sin x \cos x \tan^{- 1} (\sin x) dx$