$\int x . {cos}^{3} x^{2} . sin x^{2} d x =$

\frac{1}{8} {sin}^{4} x^{2} + c

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\frac{1}{8} {cos}^{4} x^{2} + c

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- \frac{1}{8} {cos}^{4} x^{2} + c

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- \frac{1}{8} {sin}^{4} x^{2} + c

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Solution

The correct option is D $- \frac{1}{8} {cos}^{4} x^{2} + c$
Let $x^{2} = t$

$2 x . d x = d t$

$x d x = \frac{1}{2} d t$

$I = \int x . {cos}^{3} x^{2} . sin x^{2} d x = \frac{1}{2} \int {cos}^{3} t . sin t d t$

Let $cos t = p$

$sin t d t = - d p$

$I = \frac{1}{2} \int {cos}^{3} t . sin t d t = - \frac{1}{2} \int p^{3} d p = - \frac{1}{8} p^{4} + c$

$= - \frac{1}{8} {cos}^{4} t + c = - \frac{1}{8} {cos}^{4} x^{2} + c$

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∫x.cos3x2.sinx2dx=

The correct option is D −18cos4x2+cLet x2=t2x.dx=dtxdx=12dtI=∫x.cos3x2.sinx2dx=12∫cos3t.sintdtLet cost=psintdt=−dpI=12∫cos3t.sintdt=−12∫p3dp=−18p4+c=−18cos4t+c=−18cos4x2+c

$\int x . {cos}^{3} x^{2} . sin x^{2} d x =$