Integration of trig functions: int sin^2 x cos^4 x dx

[chengeto]

$\displaystyle \int sin^2xcos^4xdx$

$\displaystyle \int(1-cos^2x)(cos^4x)$

$\displaystyle \int(cos^4x-cos^6x)dx$

$\displaystyle \cos^2 x = \frac {1 + \cos 2x}2$

$\displaystyle \frac{1}{2}\int cos^4xdx-\frac{1}{2}\int cos^6xdx$

$\displaystyle \frac{1}{8}\int (1+cos2x)^2-\frac{1}{16}\int(1+cos2x)^3$

I'm stuck here because it seems as if l am chasing my tail.

 

[galactus]

Re: Integration of trig functions

$\displaystyle \int sin^{2}(x)cos^{4}(x)dx$

$\displaystyle =\frac{1}{8}\int (1-cos(2x))(1+cos(2x))^{2}dx$

$\displaystyle =\frac{1}{8}\int (1-cos^{2}(2x))(1+cos(2x))dx$

$\displaystyle =\frac{1}{8}\int sin^{2}(2x)dx+\frac{1}{8}\int sin^{2}(2x)cos(2x)dx$

$\displaystyle =\frac{1}{16}\int (1-cos(4x))dx+\frac{1}{48}sin^{3}(2x)$

$\displaystyle \boxed{=\frac{1}{16}x-\frac{1}{64}sin(4x)+\frac{1}{48}sin^{3}(2x)+C}$

 

[chengeto]

Re: Integration of trig functions

$\displaystyle \int sin^{2}(x)cos^{4}(x)dx$

$\displaystyle =\frac{1}{8}\int (1-cos(2x))(1+cos(2x))^{2}dx$

$\displaystyle =\frac{1}{8}\int (1-cos^{2}(2x))(1+cos(2x))dx$

$\displaystyle =\frac{1}{8}\int sin^{2}(2x)dx+\frac{1}{8}\int sin^{2}(2x)cos(2x)dx$

$\displaystyle =\frac{1}{16}\int (1-cos(4x))dx+\frac{1}{48}sin^{3}(2x)$

$\displaystyle \boxed{=\frac{1}{16}x-\frac{1}{64}sin(4x)+\frac{1}{48}sin^{3}(2x)+C}$

Galactus how did you get :


\(\displaystyle \frac{1}{48}sin^{3}(2x)+C}\)

 

[galactus]

Just integrated.