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Question

Question: \(\int_{- \pi/2}^{\pi/2}{\sin^{2}x\cos^{2}x(\sin x + \cos x)dx =}\)...

∫−π/2π/2sin⁡2xcos⁡2x(sin⁡x+cos⁡x)dx=\int_{- \pi/2}^{\pi/2}{\sin^{2}x\cos^{2}x(\sin x + \cos x)dx =}

A

215\frac{2}{15}

B

415\frac{4}{15}

C

615\frac{6}{15}

D

815\frac{8}{15}

Answer

415\frac{4}{15}

Explanation

Solution

∫−π/2π/2sin⁡2xcos⁡2x(sin⁡x+cos⁡x)dx\int_{- \pi/2}^{\pi/2}{\sin^{2}x\cos^{2}x(\sin x + \cos x)dx}

=∫−π/2π/2sin⁡3xcos⁡2xdx+∫−π/2π/2sin⁡2xcos⁡3xdx\int_{- \pi/2}^{\pi/2}{\sin^{3}x\cos^{2}xdx + \int_{- \pi/2}^{\pi/2}{\sin^{2}x\cos^{3}xdx}}

=0+2∫0π/2sin⁡2xcos⁡3xdx=0+2×215=415= 0 + 2\int_{0}^{\pi/2}{\sin^{2}x\cos^{3}xdx} = 0 + 2 \times \frac{2}{15} = \frac{4}{15} .