Can u integrate this for me?

Question

sin^2(2x+5) wid respect to x , also explain the steps used.

Answer 1

Use the identity sin^2(t) = (1 + cos(2t))/2.

Answer 2

First we use this trigonometric identity:

sin^2(a) = [1-cos(2a)]/2

Or in this case:

sin^2(2x+5) = [1-cos(4x+10)]/2

now we split the identity:

1/2 - [cos(4x+10)]/2

we are ready to integrate; but for the second part of this we must find the proper differential. Then we make u =  4x+10; and du = 4dx. Now we integrate:

∫1/2 dx - ∫4[cos(4x10)]/8 dx

or:

∫1/2 dx - 1/8∫[cos(4x+10)] 4dx  

finally answer is:

x/2 - [sin(4x+10)]/8 + c  (when c is the integration constant)

Answer 3

Let's use "integration by parts" to solve this.

Let u=sin(2x+5)

     dv=sin(2x+5)dx

     du=2cos(2x+5)dx

     v=-1/2cos(2x+5)

Then

∫ sin^2(2x+5)dx = (-1/2)sin(2x+5)(cos(2x+5) + (1/2) ∫cos^2(2x+5) dx

Moving a term to the left gives

∫ sin^2(2x+5)dx - (1/2)∫ cos^2(2x+5)dx = (-1/2)sin(2x+5)cos(2x+5)

using trig identities gives

∫ sin^2(2x+5)dx - (1/2)∫[1-sin^2(2x+5)]dx = (-1/2)sin(2x+5)cos(2x+5)

∫ sin^2(2x+5)dx - (1/2)x + (1/2)∫ sin^2(2x+5)dx = (-1/2)sin(2x+5)cos(2x+5)

(3/2)∫sin^2(2x+5)dx = (1/2)x -(1/2)sin(2x+5)cos(2x+5)

∫sin^2(2x+5)dx = (2/3)[(1/2)x - (1/2)sin(2x+5)cos(2x+5)

∫sin^2(2x+5)dx = (1/3)x - (1/3)sin(2x+5)(cos(2x+5) + C  

Answer 4

Let 2x+5 =u

2 dx = du

dx = (1/2) du

The integral becomes

(1/2) ∫ sin^2(u) du

=(1/2)  Ã¢ÂˆÂ« (1-cos(2u))/2 du

=(1/4)  Ã¢ÂˆÂ«  du - (1/4)  Ã¢ÂˆÂ« cos(2u) du

(1/4) u -(1/4) sin(2u) /2

=u/4 -sin(2u)/8 +C

=(2x+5) / 4 - sin (4x+10) / 8 + C