How do you find the integral of #cos^6(x) #?

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Answer 1 · 2017-02-18T22:43:48

#int cos^6xdx = (sinxcos^5x )/6+ (sinxcos^3x)/6 + 2/5 (sinxcosx) + 2/5x +C#

Integrate by parts:

#int cos^6xdx = int cos^5x cosx dx = int cos^5x d(sinx)#

#int cos^6xdx = sinxcos^5x + 5 int cos^4x sin^2x dx#

#int cos^6xdx = sinxcos^5x + 5 int cos^4x (1 - cos^2x) dx#

#int cos^6xdx = sinxcos^5x + 5 int cos^4x - 5int cos^6x dx#

As the integral is on both sides we can solve for it:

#6 int cos^6xdx = sinxcos^5x + 5 int cos^4xdx #

#int cos^6xdx = (sinxcos^5x )/6+ 5/6 int cos^4xdx #

We can use the same method to reduce the degree of #cosx# again:

#int cos^4xdx = int cos^3x d(sinx) #

#int cos^4xdx = sinxcos^3x + 4 int cos^2x sin^2x dx #

#int cos^4xdx = sinxcos^3x + 4 int cos^2xdx -4 int cos^4xdx #

#int cos^4xdx = (sinxcos^3x)/5 + 4/5 int cos^2xdx #

And again:

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

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

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

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

Putting it all together we have:

#int cos^6xdx = (sinxcos^5x )/6+ (sinxcos^3x)/6 + 2/5 (sinxcosx) + 2/5x +C#