Question

Find the integrals of the function: $sin^3(2x+1)$

Answer 1

Answer:

Step-by-step explanation:

Concept:

$sin3A=3\;sinA-4\:sin^3A\\\\sin^3A=\frac{1}{4}[3\;sinA-sin3A]$

I have applied substitution method (change of variable) to solve this integral.

In substitution method, the given integrand(non integrable function) is changed into an integrable function by taking suitable substitution.

$I=\int{sin^3(2x+1)}\:dx\\\\Take,\\\\t=2x+1\\\\\frac{dt}{dx}=2\\\\\frac{dt}{2}=dx\\\\I=\int{sin^3t}\:\frac{dt}{2}\\\\I=\frac{1}{2}\int{sin^3t}\:dt\\\\I=\frac{1}{2}\int{\frac{1}{4}[3\;sint-sin3t]}\:dt\\\\I=\frac{1}{2}.\frac{1}{4}\int{[3\;sint-sin3t]}\:dt\\\\I=\frac{1}{8}\int{[3\;sint-sin3t]}\:dt\\\\I=\frac{1}{8}[3\:(-cost)-\frac{(-cos3t)}{3}]/+c\\\\I=\frac{1}{8}[-3cost+\frac{(cos3t)}{3}]+c\\\\I=\frac{1}{8}[-3cos(2x+1)+\frac{(cos3(2x+1))}{3}]+c$