Question

How do I integrate `int cos^6theta d theta`?

Answer 1

`1/5*(cosx)^5*sinx`+`5/24*(cosx)^4*sinx`+`5/32*sin2x+(5x)/16+C`

Explanation:

After using `int (cosx)^n*dx`=`1/n*(cosx)^(n-1)sinx`+`(n-1)/n`*`int (cosx)^(n-2)*dx` reduction formula for `n=6 and 4`

`int (cosx)^6*dx`=`1/5*(cosx)^5*sinx`+`5/6`*`int (cosx)^4*dx`

=`1/5*(cosx)^5*sinx`+`5/6`(`1/4(cosx)^4*sinx`+`3/4``int (cosx)^2*dx`)

=`1/5*(cosx)^5*sinx`+`5/24*(cosx)^4*sinx`+`5/8`*`int (cosx)^2*dx`

=`1/5*(cosx)^5*sinx`+`5/24*(cosx)^4*sinx`+`5/16`*`int(1+cos2x)*dx`

=`1/5*(cosx)^5*sinx`+`5/24*(cosx)^4*sinx`+`5/32*sin2x+(5x)/16+C`