How to integrate x(sinx)^2 ?

Question

Please tell me How to integrate x(sinx)^2

Answer 1

∫x sin²(x)dx

Let u = x

du = dx

Let dv = sin²(x) dx

v =  ÃƒÂ¢Ã‚ˆÂ«sin²(x) dx  = (1/2)∫dx - (1/2)∫cos(2x)dx = x/2 - sin(2x)/4 + c

v = x/2 - sin(2x)/4 + c

∫udv = uv - ∫vdu

∫x sin²(x)dx  = x(x/2 - sin(2x)/4) - ∫(x/2 - sin(2x)/4)dx + c

∫x sin²(x)dx  =  x²/2 - xsin(2x)/4 - x²/4 + 1/4∫sin(2x)dx + c

∫x sin²(x)dx  =  x²/2 - xsin(2x)/4 + x²/4 + 1/4(-cos(2x)/2) + c

∫x sin²(x)dx  = x²/4 - xsin(2x)/4  - cos(2x)/8 + c

Answer 2

∫ (  x sin^2(x) dx )

First, use the half angle identity

sin^2(x) = (1/2)(1 - cos(2x))

∫ (  x (1/2) (1 - cos(2x))  dx )

Factor out the constant (1/2) from the integral.

(1/2) ∫ (  x (1 - cos(2x)) dx )

Distribute the x, to get

(1/2) ∫ ( [ x - x cos(2x) ] dx )

We can split this into two integrals, to get

(1/2) [ ∫ ( x dx ) - ∫ (  x cos(2x) dx ) ]

(1/2) [ (1/2)x^2  -  Ã¢ÂˆÂ« ( x cos(2x) dx ) ]

Distribute the (1/2), to get

(1/4) x^2 - (1/2) ∫ (  x cos(2x) dx )

From here, we can use integration by parts.

Let u = x.  dv = cos(2x) dx.

du = dx.  v = (1/2) sin(2x)

Apply the integration by parts, to get

(1/4) x^2 - (1/2) [ (1/2)x sin(2x) - ∫ ( (1/2)sin(2x) dx ) ]

Distribute the (1/2), to get

(1/4) x^2 - (1/4) x sin(2x) + (1/2) ∫ (  (1/2) sin(2x) dx )

And now, factor the (1/2) from the remaining integral.  This merges with the other (1/2) to make (1/4).

(1/4) x^2 - (1/4) x sin(2x) + (1/4) ∫( sin(2x) dx )

Which is now an easy integral, as the integral of sin(2x) is equal to (-1/2)cos(2x).

(1/4) x^2 - (1/4) x sin(2x) + (1/4) (-1/2) cos(2x) + C

Simplify,

(1/4) x^2 - (1/4) x sin(2x) - (1/8)cos(2x) + C

Answer 3

∫x sin^2 x dx=

∫x(1-cos 2x)/2 dx=

1/2∫x dx - 1/2∫x cos 2x dx=

(1/2)x - 1/8∫(2x) cos(2x) d(2x)

FRom here y=2x and computing

∫y cos y dy by parts

∫y cos y dy=

∫y d sin y= y sin y - ∫sin y dy=

ysin y +cos y +C

Substitute back and OK

Answer 4

by using LATE rule

Answer 5

∫sin²x dx

= (1/2) ∫2sin²x dx

= (1/2) ∫(1 - cos 2x) dx

= x/2 - (1/4)sin 2x ...   (1)

(Cont. of integration is not shown as the result is to be used later.)

∫xsin²x dx

= x ∫sin²x dx - ∫[d/dx(x) ∫sin²x dx] dx

= x[x/2 - (1/4)sin 2x] - ∫[x/2 - (1/4)sin 2x] dx   [using result (1)]

=  x²/2 - (x/4)sin 2x - x²/4 - (1/8)cos 2x + c

= x²/4 - (x/4)sin 2x - (1/8)cos 2x + c