How do you integrate #(cosx)^2dx#?

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Answer 1 · 2016-12-01T14:09:43

#int cos^2xdx = 1/2(x+1/2sin2x) + C#

Considering that:

#(dsinx)/dx = cosx#

#(dcosx)/dx = -sinx#

#int cos^2xdx =int cosx * cosx dx =int cosx d(sinx)#

Integrating by parts:

#int cos^2xdx = sinxcosx - int sinx dcosx = #

# = sinxcosx + int sin^2x dx = #

# = sinxcosx + int (1-cos^2x) dx = #

# = sinxcosx + x - int cos^2x dx #

So:

#2int cos^2xdx = sinxcosx + x#

and finally:

#int cos^2xdx = 1/2(x+1/2sin2x) + C#

An alternative method is to use the identity:

#cos(2x) = cos^2x-sin^2x = cos^2x - (1-cos^2x) = #
# = 2cos^2x-1#

so that:

#cos^2x = (1+cos2x)/2#

#int cos^2xdx = int (1+cos2x)/2 dx = 1/2(x+1/2sin2x) + C#