Question

Find the integrals of the function: $sin^3\ x\ cos^3\ x$

Answer 1

Answer:

Step-by-step explanation:

Formula used:

$1.cos^2\theta=1-sin^2\theta\\\\2.\int{x^n}\:dx=\frac{x^{n+1}}{n+1}+c$

$Let,\\\\I=\int{sin^3x.cos^3x}\:dx\\\\I=\int{sin^3x.cos^2x.cosx}\:dx\\\\I=\int{sin^3x(1-sin^2x)cosx}\:dx\\\\I=\int{(sin^3x-sin^5x)cosx}\:dx$

Take

t=sinx

$\frac{dt}{dx}=cosx\\\\dt=cosx\:dx$

Now,

$I=\int{(t^3-t^5)}\:dt\\\\I=\frac{t^4}{4}-\frac{t^6}{6}+c\\\\I=\frac{sin^4x}{4}-\frac{sin^6x}{6}+c$