What is the antiderivative of x*(sinx)^2x(sinx)2?

1 Answer
May 31, 2018

1/4(x^2 -xsin2x -1/2 cos 2x) + C14(x2xsin2x12cos2x)+C

Explanation:

We have to calculate:-

intx(sin x)^2dxx(sinx)2dx.

= int xsin^2xdx=xsin2xdx

As, We know, cos 2x = 1 - 2sin^2xcos2x=12sin2x,

Then 2sin^2x = 1 - cos 2x2sin2x=1cos2x

rArr sin^2 x = (1 - cos 2x)/2sin2x=1cos2x2....................(i)

Now, Our integral Turns into

int x ((1 - cos 2x)/2)dxx(1cos2x2)dx

= int x/2 (1 - cos 2x) dx=x2(1cos2x)dx

Now We will use Integration by Parts.

So, int x/2 (1 - cos 2x)dxx2(1cos2x)dx

= x/2 int (1 - cos 2x)dx - int(d/dx(x/2)int (1 - cos 2x))dx=x2(1cos2x)dx(ddx(x2)(1cos2x))dx

= x/2 (int dx - int cos2xdx) - int(1/2(intdx - int cos 2x dx)dx=x2(dxcos2xdx)(12(dxcos2xdx)dx............(ii)

Now, Lets solve int cos2xdxcos2xdx.

Substitute u = 2xu=2x.

So, du = 2dx rArr dx = 1/2 dudu=2dxdx=12du

So, int cos2xdxcos2xdx

= 1/2intcos u du=12cosudu

= 1/2sinu + C=12sinu+C

= 1/2sin2x + C=12sin2x+C

So, From (i),

We have,
= x/4 (2x - sin2x) - 1/4int(2x - sin2x)dx=x4(2xsin2x)14(2xsin2x)dx

= x/4(2x - sin2x) - 1/4 xx 2intxdx + 1/4int sin 2xdx=x4(2xsin2x)14×2xdx+14sin2xdx

= x/4(2x - sin2x) - 1/4x^2 + 1/4int sin 2xdx=x4(2xsin2x)14x2+14sin2xdx

= 1/2x^2 - 1/4xsin2x - 1/4x^2 + 1/4intsin2xdx=12x214xsin2x14x2+14sin2xdx.................(iii)

Now,

intsin2xdxsin2xdx

Substitute u = 2x rArr du = 2dx rArr dx = 1/2duu=2xdu=2dxdx=12du

So,

intsin2xdxsin2xdx

= 1/2intsinudu=12sinudu

= -1/2cosu + C=12cosu+C

= -1/2cos 2x + C=12cos2x+C

So, The Final Integration (From (iii)),

= 1/4x^2 - 1/4xsin2x - 1/8cos 2x + C=14x214xsin2x18cos2x+C

= 1/4(x^2 - xsin2x - 1/2cos 2x) + C=14(x2xsin2x12cos2x)+C

Hop ethis helps.