∫0π sin3⁡x1+cos2⁡xdx=

∫0π sin3⁡x1+cos2⁡xdx=∫0π sin2⁡x⋅sin⁡x1+cos2⁡xdx=∫0π 1−cos2⁡x1+cos2⁡x⋅sin⁡xdx Put cos⁡x=t. Then −sin⁡x⋅dx=dt x=0,π⇒t=1,−1 ∫0π sin3⁡x1+cos2⁡xdx=∫1−1 1−t21+t2(−dt)=∫1−1 t2−1t2+1dt=∫1−1 t2t2+1dt−∫1−1 1t2+1dt=∫1−1 1+t2−11+t2dt−∫1−1 11+t2dt =∫1−1 1+t21+t2dt−∫1−1 11+t2dt−∫1−1 11+t2dt=∫1−1 dt−2∫1−1 11+t2dt=[t]1−1−2Tan−1⁡t1−1 =[−1−1]−2Tan−1⁡(−1)−Tan−1⁡(1)=−2−2−π4−π4=−2−2−π2=π−2

∫0π sin3⁡x1+cos2⁡xdx=

∫0π sin3⁡x1+cos2⁡xdx=∫0π sin2⁡x⋅sin⁡x1+cos2⁡xdx=∫0π 1−cos2⁡x1+cos2⁡x⋅sin⁡xdx Put cos⁡x=t. Then −sin⁡x⋅dx=dt x=0,π⇒t=1,−1 ∫0π sin3⁡x1+cos2⁡xdx=∫1−1 1−t21+t2(−dt)=∫1−1 t2−1t2+1dt=∫1−1 t2t2+1dt−∫1−1 1t2+1dt=∫1−1 1+t2−11+t2dt−∫1−1 11+t2dt =∫1−1 1+t21+t2dt−∫1−1 11+t2dt−∫1−1 11+t2dt=∫1−1 dt−2∫1−1 11+t2dt=[t]1−1−2Tan−1⁡t1−1 =[−1−1]−2Tan−1⁡(−1)−Tan−1⁡(1)=−2−2−π4−π4=−2−2−π2=π−2

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$\int_{0}^{π} \frac{\sin^{3} x}{1 + \cos^{2} x} dx =$

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$π - 1$

$π + 2$

$π - 2$

$2 π - 2$

answer is C.

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Detailed Solution

$\int_{0}^{π} \frac{\sin^{3} x}{1 + \cos^{2} x} dx = \int_{0}^{π} \frac{\sin^{2} x \cdot \sin x}{1 + \cos^{2} x} dx = \int_{0}^{π} \frac{1 - \cos^{2} x}{1 + \cos^{2} x} \cdot \sin xdx Put \cos x = t . Then - \sin x \cdot dx = dt x = 0, π \Rightarrow t = 1, - 1 \begin{array}{l} \int_{0}^{π} \frac{\sin^{3} x}{1 + \cos^{2} x} dx = \int_{1}^{- 1} \frac{1 - t^{2}}{1 + t^{2}} (- dt) = \int_{1}^{- 1} \frac{t^{2} - 1}{t^{2} + 1} dt = \int_{1}^{- 1} \frac{t^{2}}{t^{2} + 1} dt - \int_{1}^{- 1} \frac{1}{t^{2} + 1} dt = \int_{1}^{- 1} \frac{1 + t^{2} - 1}{1 + t^{2}} dt - \int_{1}^{- 1} \frac{1}{1 + t^{2}} dt \\ = \int_{1}^{- 1} \frac{1 + t^{2}}{1 + t^{2}} dt - \int_{1}^{- 1} \frac{1}{1 + t^{2}} dt - \int_{1}^{- 1} \frac{1}{1 + t^{2}} dt = \int_{1}^{- 1} dt - 2 \int_{1}^{- 1} \frac{1}{1 + t^{2}} dt = [t]_{1}^{- 1} - 2 {[{Tan}^{- 1} t]}_{1}^{- 1} \\ = [- 1 - 1] - 2 [{Tan}^{- 1} (- 1) - {Tan}^{- 1} (1)] = - 2 - 2 [- \frac{π}{4} - \frac{π}{4}] = - 2 - 2 [- \frac{π}{2}] = π - 2 \end{array}$