How do you integrate $int cos^2x$ by integration by parts method?

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Answer 1 · 2017-01-23T10:42:25

$x/2+1/2sin x cos x + c$

If you really want to integrate by parts, choose $u=cos x$ , $dv= cos x dv$ , $du=-sin xdx$ , $v = sin x$ .

$int udv = uv - int v du$

$int cosx cosx dx= cos x sinx - int sin x (-sin x)dx$

$int cos^2 x dx= cos x sin x + int (1 - cos^2x)dx$

$int cos^2 x dx= cos x sin x + int 1 dx - int cos^2x dx$

Now for the sneaky part: take the integral on the right over to the left:

$2int cos^2x dx = cos x sin x + x$

Hence
$int cos^2xdx = 1/2 x + 1/2 sin x cos x$

However, a shorter way is to use the identities $cos2x = cos^2x-sin^2x = 2 cos^2 x - 1 = 1 - 2sin^2 x$ and $sin2x=2sinxcosx$ .

$int cos^2 x=int (1+cos2x)/2dx$

$=int1/2 dx + 1/2 int cos2x dx$

$=1/2x +1/2sin 2x+c$

$=1/2x+sinxcosx+c$

How do you integrate int cos^2xint cos^2x by integration by parts method?