- x cos 3x2 sinx2 d x is equal toA- -cos 4x2 sinx2-16-CB- cosx2-16-CC- -cos 4x2-8-CD- cos 3x4-8-C

The correct option is C cos^4 (x^2)8+C∫ xcos ^3 (x^2)sin (x^2)dx nbsp; Let x^2=t⇒ 2x dx=dt =12∫cos ^3 (t)sin (t) dt Let cos t=z⇒ sin t dt=dz = 12∫ z ^3 dx= ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

∫ x cos 3x2 s...

Question

$\int x {cos}^{3} (x^{2}) sin (x^{2}) d x$ is equal to

- \frac{{cos}^{4} (x^{2}) sin (x^{2})}{16} + C

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\frac{cos (x^{2})}{16} + C

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- \frac{{cos}^{4} (x^{2})}{8} + C

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\frac{{cos}^{3} (x^{4})}{8} + C

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Solution

The correct option is C $- \frac{{cos}^{4} (x^{2})}{8} + C$
$\int x {cos}^{3} (x^{2}) sin (x^{2}) d x$
Let $x^{2} = t \Rightarrow 2 x d x = d t$
$= \frac{1}{2} \int {cos}^{3} (t) sin (t) d t$
Let $cos t = z \Rightarrow - sin t d t = d z$
$= - \frac{1}{2} \int z^{3} d x = - \frac{z^{4}}{8} + c_{1} = - \frac{{cos}^{4} (x^{2})}{8} + C$

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∫xcos3(x2)sin(x2)dx is equal to

The correct option is C −cos4(x2)8+C∫xcos3(x2)sin(x2)dx Let x2=t⇒2x dx=dt =12∫cos3(t)sin(t) dt Let cost=z⇒−sint dt=dz =−12∫z3dx=−z48+c1=−cos4(x2)8+C

$\int x {cos}^{3} (x^{2}) sin (x^{2}) d x$ is equal to