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# Indefinite Integral......

0
1678
4

I thank you for any help.

Jan 1, 2016

#1
+111321
+15

This type of integral isn't that difficult.....just tedious.....!!!!!

∫(sin(x))^4   dx  =

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

∫  [ 1/2  - cos(2x)/2]^2 dx  =

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

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

For the second integral    let  u =  2x        du  = 2 dx   du /2  = dx

[1/4]x - (1/2) ∫ cos(u)  du/2   + (1/4) ∫  cos^2(2x)  dx

[1/4]x - (1/4) ∫ cos(u)  du   + (1/4) ∫  cos^2(2x)  dx

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

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

For the last integral     ........  cos^2(2x)  =  [ 1 + cos(4x] / 2   =  1/2 + cos(4x)/2

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

[1/4]x - [1/4]sin(2x) + [1/8]x   + (1/8) ∫ cos(4x)  dx

[3/8]x  - [1/4]sin(2x) + (1/8) ∫ cos(4x)  dx

As before.....let u = 4x       du  = 4dx       du/4 = dx

[3/8]x -[1/4]sin(2x) + (1/8)  ∫ cos(u) du/4

[3/8]x -[1/4]sin(2x) + (1/32)  ∫ cos(u) du

[3/8]x -[1/4]sin(2x) + (1/32) sin(u)  + C =

[3/8]x -[1/4]sin(2x) + (1/32) sin(4x) + C =

[1/32] [ 12x - 8sin(2x) + sin(4x) ] + C

Jan 1, 2016
edited by CPhill  Jan 1, 2016

#1
+111321
+15

This type of integral isn't that difficult.....just tedious.....!!!!!

∫(sin(x))^4   dx  =

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

∫  [ 1/2  - cos(2x)/2]^2 dx  =

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

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

For the second integral    let  u =  2x        du  = 2 dx   du /2  = dx

[1/4]x - (1/2) ∫ cos(u)  du/2   + (1/4) ∫  cos^2(2x)  dx

[1/4]x - (1/4) ∫ cos(u)  du   + (1/4) ∫  cos^2(2x)  dx

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

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

For the last integral     ........  cos^2(2x)  =  [ 1 + cos(4x] / 2   =  1/2 + cos(4x)/2

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

[1/4]x - [1/4]sin(2x) + [1/8]x   + (1/8) ∫ cos(4x)  dx

[3/8]x  - [1/4]sin(2x) + (1/8) ∫ cos(4x)  dx

As before.....let u = 4x       du  = 4dx       du/4 = dx

[3/8]x -[1/4]sin(2x) + (1/8)  ∫ cos(u) du/4

[3/8]x -[1/4]sin(2x) + (1/32)  ∫ cos(u) du

[3/8]x -[1/4]sin(2x) + (1/32) sin(u)  + C =

[3/8]x -[1/4]sin(2x) + (1/32) sin(4x) + C =

[1/32] [ 12x - 8sin(2x) + sin(4x) ] + C

CPhill Jan 1, 2016
edited by CPhill  Jan 1, 2016
#2
+5

Thank you very much CPhill. Sorry to have bothered you, on New Year's day no less!.

Jan 1, 2016
#3
+10

Take the integral:
integral sin^4(x) dx
Use the reduction formula,  integral sin^m(x) dx = -(cos(x) sin^(m-1)(x))/m + (m-1)/m integral sin^(-2+m)(x) dx, where m = 4:
=  -1/4 sin^3(x) cos(x)+3/4 integral sin^2(x) dx
Write sin^2(x) as 1/2-1/2 cos(2 x):
=  -1/4 sin^3(x) cos(x)+3/4 integral (1/2-1/2 cos(2 x)) dx
Integrate the sum term by term and factor out constants:
=  -1/4 sin^3(x) cos(x)-3/8 integral cos(2 x) dx+3/8 integral 1 dx
For the integrand cos(2 x), substitute u = 2 x and  du = 2  dx:
=  -1/4 sin^3(x) cos(x)-3/16 integral cos(u) du+3/8 integral 1 dx
The integral of cos(u) is sin(u):
=  -(3 sin(u))/16-1/4 sin^3(x) cos(x)+3/8 integral 1 dx
The integral of 1 is x:
=  -(3 sin(u))/16+(3 x)/8-1/4 sin^3(x) cos(x)+constant
Substitute back for u = 2 x:
=  (3 x)/8-1/4 sin^3(x) cos(x)-3/8 sin(x) cos(x)+constant
Which is equal to:
Answer: | =  1/32 (12 x-8 sin(2 x)+sin(4 x))+constant

Jan 2, 2016
#4
+109518
+5

Hi, thanks CPhill and guest.

I am just going to look at this bit

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

$$sin^2x =\frac{ 1 - cos(2x)}{2}$$

I am hopeless (also stuborn) aboul learning formulas.  I just remember what I feel i have to remember and work the others out each time.

So I do not remember this one :/

$$Cos(A+B)=cosAcosB-sinAsinB\qquad \mbox{I have committed this to memory}\\ cos(2x)=cos^2x-sin^2x\\ cos(2x)=1-sin^2x\;\;\;\;\;-sin^2x\\ cos(2x)=1-2sin^2x\\ cos(2x)-1=-2sin^2x\\ \frac{cos(2x)-1}{-2}=sin^2x\\ sin^2x=\frac{1-cos(2x)}{2}\\$$

If I need sin^2(2x)   or  cos^2(x)  etc I got through a similar procedure.

It doesn't take me long because I have had a LOT of practice :)

Jan 2, 2016