∫0π/2 sin3⁡xcos⁡xdxsin4⁡x+cos4⁡x=

Let I=∫0π/2 sin3⁡xcos⁡xsin4⁡x+cos4⁡xdx→(1)⇒I=∫0π/2 sin3⁡(π/2−x)cossin4⁡(π/2−x)+c4⋅s4π/2−x)(π/2−x)dx ⇒I=∫0π/2 cos3⁡xsin⁡xcos4⁡x+sin4⁡xdx→(2)2I=∫0π/2 sin3⁡xcos⁡x+cos3⁡xsinx sin4⁡x+cos4⁡xdx =∫0π/2 sin⁡xcos⁡xsin2⁡x+cos2⁡xsin4⁡x+cos4⁡xdx⇒2I=∫0π/2 sin⁡xcos⁡xsin4⁡x+cos4⁡xdx=∫0π/2 sec2⁡xtan⁡x1+tan4⁡xdx Put tan2⁡x=t. Then 2tan⁡xsec2⁡xdx=dtL.L & U.L ⇒x=0,π/2⇒t=0,∞⇒2I=12∫0∞ 11+t2dt=12Tan−1⁡t0∞=12π2−0=π4⇒2I=π4⇒I=π8.

∫0π/2 sin3⁡xcos⁡xdxsin4⁡x+cos4⁡x=

Let I=∫0π/2 sin3⁡xcos⁡xsin4⁡x+cos4⁡xdx→(1)⇒I=∫0π/2 sin3⁡(π/2−x)cossin4⁡(π/2−x)+c4⋅s4π/2−x)(π/2−x)dx ⇒I=∫0π/2 cos3⁡xsin⁡xcos4⁡x+sin4⁡xdx→(2)2I=∫0π/2 sin3⁡xcos⁡x+cos3⁡xsinx sin4⁡x+cos4⁡xdx =∫0π/2 sin⁡xcos⁡xsin2⁡x+cos2⁡xsin4⁡x+cos4⁡xdx⇒2I=∫0π/2 sin⁡xcos⁡xsin4⁡x+cos4⁡xdx=∫0π/2 sec2⁡xtan⁡x1+tan4⁡xdx Put tan2⁡x=t. Then 2tan⁡xsec2⁡xdx=dtL.L & U.L ⇒x=0,π/2⇒t=0,∞⇒2I=12∫0∞ 11+t2dt=12Tan−1⁡t0∞=12π2−0=π4⇒2I=π4⇒I=π8.

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$\int_{0}^{π / 2} \frac{\sin^{3} x \cos x d x}{\sin^{4} x + \cos^{4} x} =$

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$π$

$π / 2$

$π / 4$

$π / 8$

answer is D.

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Detailed Solution

Let $I = \int_{0}^{π / 2} \frac{\sin^{3} x \cos x}{\sin^{4} x + \cos^{4} x} d x \to (1)$

$\begin{array}{l} \Rightarrow I = \int_{0}^{π / 2} \frac{\sin^{3} (π / 2 - x) \cos}{\sin^{4} (π / 2 - x) + c^{4} \cdot s^{4}} \frac{π / 2 - x)}{(π / 2 - x)} d x \Rightarrow I = \int_{0}^{π / 2} \frac{\cos^{3} x \sin x}{\cos^{4} x + \sin^{4} x} d x \to (2) \\ 2 I = \int_{0}^{π / 2} \frac{\sin^{3} x \cos x + \cos^{3} x \sin x^{}}{\sin^{4} x + \cos^{4} x} d x = \int_{0}^{π / 2} \frac{\sin x \cos x (\sin^{2} x + \cos^{2} x)}{\sin^{4} x + \cos^{4} x} d x \end{array}$

$\begin{array}{l} \Rightarrow 2 I = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \int_{0}^{π / 2} \frac{\sec^{2} x \tan x}{1 + \tan^{4} x} d x \\ Put \tan^{2} x = t . Then 2 \tan x \sec^{2} x d x = d t \\ L . L & U . L \Rightarrow x = 0, π / 2 \Rightarrow t = 0, \infty \\ \Rightarrow 2 I = \frac{1}{2} \int_{0}^{\infty} \frac{1}{1 + t^{2}} d t = \frac{1}{2} {({Tan}^{- 1} t)}_{0}^{\infty} = \frac{1}{2} (\frac{π}{2} - 0) = \frac{π}{4} \\ \Rightarrow 2 I = \frac{π}{4} \Rightarrow I = \frac{π}{8} . \end{array}$