How do you find the integral of $x * cos^2 (x)dx$ ?

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Answer 1 · 2018-04-27T08:20:03

$I=x^2/4+1/8[2xsin2x+cos2x]+c$

We know that,

$color(red)((1)cos^2theta=(1+cos2theta)/2$

$color(blue)((2)intcosAxdx=(sinAx)/A+c$

$color(violet)((3)intsinAxdx=(-cosAx)/A+c$

Here,

$I=intxcos^2xdx$

$=intx((1+cos2x)/2)dx...toApply(1)$

$=1/2intxdx+1/2intxcos2xdx$

$"Using "color(blue)"Integration by Parts"$ in second integral

$I=1/2x^2/2+1/2[x int cos2xdx-intd/(dx)(x)intcos2xdx)dx]$

$=x^2/4+1/2[x((sin2x)/2)-int(1)((sin2x)/2)dx]...toApply(2)$

$=x^2/4+1/4[xsin2x-intsin2xdx]$

$=x^2/4+1/4[xsin2x-((-cos2x)/2)]+c...toApply(3)$

$I=x^2/4+1/8[2xsin2x+cos2x]+c$

How do you find the integral of x * cos^2 (x)dxx * cos^2 (x)dx?