A) \[\frac{2}{15}\]
B) \[\frac{4}{15}\]
C) \[\frac{6}{15}\]
D) \[\frac{8}{15}\]
Correct Answer: B
Solution :
\[\int_{-\pi /2}^{\pi /2}{{{\sin }^{2}}x{{\cos }^{2}}x(\sin x+\cos x)dx}\] =\[\int_{-\pi /2}^{\pi /2}{{{\sin }^{3}}x{{\cos }^{2}}xdx+\int_{-\pi /2}^{\pi /2}{{{\sin }^{2}}x{{\cos }^{3}}x\,dx}}\] \[=0+2\int_{0}^{\pi /2}{{{\sin }^{2}}x{{\cos }^{3}}xdx}\]\[=0+2\times \frac{2}{15}=\frac{4}{15}\] .
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