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How do you find the integral of # cos^4 (x) dx#?

1 Answer
Feb 28, 2017

Answer:

# int cos^4xdx = (sinxcos^3x)/4 + 3/8(cosxsinx) + 3/8x#

Explanation:

Write: #cos^4x = cos^3x*cosx# and integrate by parts:

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

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

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

Now use the identity:

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

#int cos^4xdx = sinxcos^3x + 3int (1-cos^2x) cos^2xdx#

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

We have now the same integral on both sides and we can solve for it:

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

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

Using the same process:

#int cos^2x dx = int cosxd(sinx) = cosxsinx + int sin^2xdx#

#int cos^2x dx = int cosxd(sinx) = cosxsinx + int (1-cos^2x)dx#

#int cos^2x dx = int cosxd(sinx) = cosxsinx + x - int cos^2xdx#

#int cos^2x dx = (cosxsinx)/2 + x/2#

Substituting in the expression above:

# int cos^4xdx = (sinxcos^3x)/4 + 3/8(cosxsinx) + 3/8x#