Question

integrate 6cos t/(2+sin t)^3 dt

Answer 1

Answer:

$\int{\frac{6\:cost}{(2+sint)^3}}\:dt=\frac{-3}{(2+sint)^2}+C$

Explanation:

Formula:

$\int{x^n}\:dx=\frac{x^{n+1}}{n+1}+c$

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

Let

$I=\int{\frac{6\:cost}{(2+sint)^3}}\:dt$

Take,

$x=2+sint$

$\frac{dx}{dt}=cost$

$dx=cost\:dt$

Now,

$I=6\int{\frac{1}{x^3}}\:dx$

$I=6\int{x^{-3}}\:dx$

$I=6(\frac{x^{-2}}{-2})+C$

$I=-3x^{-2}+C$

$I=\frac{-3}{x^2}+C$

$I=\frac{-3}{(2+sint)^2}+C$