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

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Answer 1 · 2017-02-28T18:53:36

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

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:

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

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