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## Integral Calculus (2017 edition)

### Unit 6: Lesson 5

Integration using trigonometric identities

# Integral of cos^3(x)

A specialized form of u-substitution involves taking advantage of trigonometric identities.

## Want to join the conversation?

• Why not substitute "sinx = u" right at the beginning? Without having to distribute the brackets.
From ∫cosx*(1-sin^2 x) dx you will get:
∫(1-u^2)du = ∫du - ∫u^2 du = u - (1/3)*u^3
Substitute back and i got the same result:
sinx-((1/3)*sin^3 x) + C
• It's valid as well, from what I can tell. Of course, with trig many times, there are multiple ways to work it: Sal seems to prefer expanding his trig functions. shrug Do as you will.
• Why doesnt u substitution work in this case in the original equation? let u = cosx
u^3 's antiderrivative is u^4 / 4, then sub cosx back in?
• That doesn't work because you do not have a du to go with the u.
If u = cos x, then du = - sin x dx
You don't have the - sin x, so you cannot make this substitution.

Remember that in integrals, to use one of the standard forms, you need to have "du" which is the derivative of whatever you decide to call u. The "du" in the notation is not just a notational requirement, it really does have to be there or you don't get the correct answer.
If the problem had been ∫ −cos³(x) sin(x) dx then that would be as you suggested ¼ cos⁴x + C
• intg [ 1/(1+cosx) ] dx ? how will we solve it?
• There are quite a few ways to approach this.
I would use the double angle identity to rewrite cos(x) as 2cos²(x/2)-1, then let u=x/2 so dx=2du.
I'll let you take it from there . . . .
• Is there a reason that you can't use 1-sinx as u?
• Well you have a 1 - sin^2 in the integral after separating the problem into cos * cos^2.
You can try to make u = 1 - sin^2
but I don't believe that would help.
Then du = -2 sin cos
and you don't have that in the expression.

Hope I understood your question correctly.
Let me know if you meant something else.
• what is the integration for sin^4 2x
(1 vote)
• Time to break out the double angle identity substitutions:

Let:

  f(x) = sin^4(2•x)  F(x) = ∫( sin^4(2•x) )dx

For reference:

Angle Sum and Difference Identities:

 sin(α + β) = sin(α)•cos(β) + cos(α)•sin(β) sin(α – β) = sin(α)•cos(β) – cos(α)•sin(β) cos(α + β) = cos(α)•cos(β) – sin(α)•sin(β) cos(α – β) = cos(α)•cos(β) + sin(α)•sin(β)   tan(α + β) = (tan(α) + tan(β)) ÷ (1 - tan(α)•tan(β)) tan(α - β) = (tan(α) - tan(β)) ÷ (1 + tan(α)•tan(β))

Decompose trig^n(x) to equivalent trig^2(x) multiplications:

  F(x) = ∫( sin^2(2•x)•sin^2(2•x) )dx

Evaluate sub:

    u = 2•x    ( d/dx(2•x) )dx = du    ( 2•d/dx(x) )dx = du    ( 2•dx/dx )dx = du    ( 2 )dx = du    dx = ( 1/2 )du

Input sub:

  F(u) = ∫( sin^2(u)•1/2•sin^2(u)•1/2 )du  F(u) = 1/4•∫( sin^2(u)•sin^2(u) )du

Find sum angle idendity for sin^2(x) if α = β:

    cos(α + β) = cos(α)•cos(β) – >sin(α)•sin(β)<    α = β = θ    cos(θ + θ) = cos(θ)•cos(θ) – sin(θ)•sin(θ)    cos(2•θ) = cos^2(θ) – sin^2(θ)    sin^2(θ) + cos(2•θ) = cos^2(θ)    sin^2(θ) = cos^2(θ) - cos(2•θ)

Isolate equivalent for cos^2(θ) from pythagorean inedtity:

      sin^2(θ) + cos^2(θ) = 1      cos^2(θ) = 1 - sin^2(θ)    sin^2(θ) = 1 - sin^2(θ) - cos(2•θ)    sin^2(θ) + sin^2(θ) = 1 - cos(2•θ)    2•sin^2(θ) = 1 - cos(2•θ)    sin^2(θ) = 1/2 - cos(2•θ)/2    u = θ    sin^2(u) = 1/2 - cos(2•u)/2

Input trig equivalent:

  F(u) = 1/4•∫( (1/2 - cos(2•u)/2)•(1/2 - cos(2•u)/2) )du  F(u) = 1/8•∫( (1 - cos(2•u))•(1 - cos(2•u)) )du  F(u) = 1/8•∫( 1 - 2•cos(2•u) + cos^2(2•u) )du  F(u) = 1/8•( ∫(1)du - ∫(2•cos(2•u))du + ∫(cos^2(2•u))du )

Evaluate sub:

    v = 2•u    ( d/du(2•u) )du = dv    ( 2•d/dx(u) )du = dv    ( 2•du/du )du = dv    ( 2 )du = dv    du = ( 1/2 )dv

Input sub:

  F( ) = 1/8•( ∫(1)du - ∫(2•cos(v)•1/2)dv + ∫(cos^2(v)•1/2)dv )  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/2•∫(cos^2(v))dv )

Find sum angle idendity for cos^2(x) if α = β:

    cos(α + β) = >cos(α)•cos(β)< – sin(α)•sin(β)    α = β = θ    cos(θ + θ) = cos(θ)•cos(θ) – sin(θ)•sin(θ)    cos(2•θ) = cos^2(θ) – sin^2(θ)    cos^2(θ) – sin^2(θ) = cos(2•θ)    cos^2(θ) = cos(2•θ) + sin^2(θ)

Isolate equivalent for sin^2(θ) from pythagorean inedtity:

      sin^2(θ) + cos^2(θ) = 1      sin^2(θ) = 1 - cos^2(θ)    cos^2(θ) = cos(2•θ) + 1 - cos^2(θ)    cos^2(θ) + cos^2(θ) = cos(2•θ) + 1    2•cos^2(θ) = cos(2•θ) + 1    cos^2(θ) = cos(2•θ)/2 + 1/2    v = θ    cos^2(v) = cos(2•v)/2 + 1/2

Input trig equivalent:

  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/2•∫( cos(2•v)/2 + 1/2 )dv )  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•∫( cos(2•v) + 1 )dv )  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( ∫(cos(2•v))dv + ∫(1)dv ) )  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( ∫(cos(2•v))dv + v ) )

Evaluate sub:

    w = 2•v    ( d/dv(2•v) )dv = dw    ( 2•d/dv(v) )dv = dw    ( 2•dv/dv )dv = dw    ( 2 )dv = dw    dv = ( 1/2 )dw

Input sub:

  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( ∫(cos(w)•1/2)dv + v ) )  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( 1/2•∫(cos(w))dv + v ) )  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( 1/2•sin(w) + v ) )  F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/8•sin(w) + v/4 )  F( ) = 1/8•( ∫(1)du - sin(v) + 1/8•sin(w) + v/4 )  F( ) = 1/8•( u - sin(v) + 1/8•sin(w) + v/4 )  F( ) = u/8 - sin(v)/8 + 1/64•sin(w) + v/32

Back sub w → v:

    w = 2•v  F( ) = u/8 - sin(v)/8 + 1/64•sin(2•v) + v/32

Back sub v → u:

    v = 2•u  F(u) = u/8 - sin(2•u)/8 + 1/64•sin(2•2•u) + 2•u/32  F(u) = u/8 - sin(2•u)/8 + 1/64•sin(4•u) + u/16  F(u) = 2•u/16 + u/16 - sin(2•u)/8 + 1/64•sin(4•u)  F(u) = 3•u/16 - sin(2•u)/8 + 1/64•sin(4•u)

Back sub u → x:

    u = 2•x  F(x) = 3•2•x/16 - sin(2•2•x)/8 + 1/64•sin(4•2•x)  F(x) = [3•x/8 - sin(4•x)/8 + sin(8•x)/64 + C]
• How can we solve for example cos^7x ?
• Can you integrate the log of a trig function, such as log (sin x), or log cos x, without the provision of "limits".
Or does the solution necessarily require "limits", such as classic textbook problem " integration of log(sin x).dx with limits from 0 to (pi/2)"
• Yes, but it is above this level of study. You have to use the polylogarithm function.
• When I tried on my own I got this:
 ∫cos³(x)dx = ∫ cos(x) - ∫cos(x) sin(x))ˆ2 dx
Then I decided to work on ∫cos(x) sin(x))ˆ2 dx using integration by parts:
> ∫cos(x) sin(x))ˆ2 dx = (sin(x))ˆ2cos(x)-2∫cos(x) sin(x))ˆ2dx
> 3∫cos(x) sin(x))ˆ2 dx = (sin(x))ˆ2cos(x)
> ∫cos(x) sin(x))ˆ2 dx = [ (sin(x))ˆ2cos(x) ]/3
But when i substitute it at the original formula doesn't seem right.
Where is the error?
(1 vote)
• The mistake was in the setup of your functions f, f', g and g'.

sin²(x)⋅cos(x)-2⋅∫cos(x)⋅sin²(x)dx

The first part is f⋅g and within the integral it must be ∫f'⋅g. The g in the integral is ok, but the derivative of f, sin²(x), is not 2⋅sin²(x) (at least, that seems to be).

Here is you can see how ∫cos(x)⋅sin²(x) can be figured out using integration by parts:

http://s17.postimg.org/dw8h9i1nj/image.png