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Question

Question: \[\int_{}^{}{\sin^{3}xco}s^{2}xdx\]...

∫sin⁡3xcos2xdx\int_{}^{}{\sin^{3}xco}s^{2}xdx

A

cos⁡2x5−cos⁡3x3+c\frac{\cos^{2}x}{5} - \frac{\cos^{3}x}{3} + c

B

cos⁡55+cos⁡33+c\frac{\cos^{5}}{5} + \frac{\cos^{3}}{3} + c

C

sin⁡55−sin⁡33+c\frac{\sin^{5}}{5} - \frac{\sin^{3}}{3} + c

D

sin⁡55+sin⁡33+c\frac{\sin^{5}}{5} + \frac{\sin^{3}}{3} + c

Answer

cos⁡2x5−cos⁡3x3+c\frac{\cos^{2}x}{5} - \frac{\cos^{3}x}{3} + c

Explanation

Solution

I=∫sin⁡x(1−cos⁡2x)cos⁡2xdxI = \int_{}^{}{\sin x(1 - \cos^{2}x)\cos^{2}xdx}

Putcos⁡x=t⇒−sin⁡xdx=dt\cos x = t \Rightarrow - \sin xdx = dt ⇒I=−∫(t2−t4)dt=t55−t33+c=cos⁡5x5−cos⁡3x3+c.\Rightarrow I = - \int_{}^{}{(t^{2} - t^{4})dt = \frac{t^{5}}{5} - \frac{t^{3}}{3} + c = \frac{\cos^{5}x}{5} - \frac{\cos^{3}x}{3}} + c.