∫(cos^3 x)^2 dx
= ∫cos^6 x dx
Now, just as cos(6x) can be expanded into a polynomial in cos(x), cos^6(x) can be expanded into a sum of cosines of multiple angles:
= ∫(1/32)(15cos(2x)+8cos(4x)+cos(6x)+10) dx
= 1/32 (15/2 sin(2x) + 2sin(4x) + 1/6 sin(6x) + 10x) + C
Or, you can use the power reduction formula as shown here
http://www.math-prof.com/Calculus_2/Calc_Ch_07.asp
which are derived using integration by parts:
∫cos^6 x dx
u = cos^5 x, du = -5cos^4 x dx
dv = cosx dx, v = sinx
∫cos^6 x dx = cos^5x sinx + 5∫cos^4x sinx dx
Now it's easy, since you have
u = cosx
du = -sinx dx
Solve it showing detail steps..please.
Integrate : ∫(cos^3 x)^2 dx
2 answers
oops. du = -5cos^4x sinx dx
things are a bit more complicated than I showed here, since you wind up with a sin^2(x), which has to be converted into (1-cos^2 x) and you go around again.
things are a bit more complicated than I showed here, since you wind up with a sin^2(x), which has to be converted into (1-cos^2 x) and you go around again.