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Course: Calculus 2 > Unit 2
Lesson 3: Integrating using trigonometric identitiesIntegral of cos^3(x)
A specialized form of u-substitution involves taking advantage of trigonometric identities.
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Why not substitute "sinx = u" right at the beginning? Without having to distribute the brackets.
From∫cosx*(1-sin^2 x) dxyou 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(56 votes)
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.(17 votes)
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?(8 votes)
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) dxthen that would be as you suggested¼ cos⁴x + C(21 votes)
intg [ 1/(1+cosx) ] dx ? how will we solve it?(5 votes)- 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 . . . .(6 votes)
Is there a reason that you can't use 1-sinx as u?(4 votes)
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.(7 votes)
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](11 votes)
- How can we solve for example cos^7x ?(6 votes)
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)"(3 votes)- Yes, but it is above this level of study. You have to use the polylogarithm function.(3 votes)
Is the ability to spot that trig identity just the result of tons of drill-work? I am terrible at spotting those.(3 votes)- I tried multiplying and dividing the integral with -sinx so what i did wouldn't change the value of the inside. I engineered it to look like: -1/sinx*∫-sinx*(cosx)^3*dx which resulted: (-1/sinx)*(cosx)^4/4. Why didn't this work?(2 votes)
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 functionsf,f',gandg'.sin²(x)⋅cos(x)-2⋅∫cos(x)⋅sin²(x)dx
The first part isf⋅gand within the integral it must be∫f'⋅g. Thegin the integral is ok, but the derivative off,sin²(x), is not2⋅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(3 votes)
Video transcript
- [Voiceover] Let's see if we can take the indefinite integral of
cosine of X to the third power. I encourage you to pause the video and see if you can figure
this out on your own. You have given it a go and
you might have gotten stuck. Some of you all might have
been able to figure it out, but some of you all
might have gotten stuck. You're like, "Okay,
cosine to the third power. "Well, gee, if I only had a
derivative of cosine here, "if I had a negative sign
of X or a sin of X here, "maybe I could've used U substitution, "but how do I take the anti-derivative "of cosine of X to the third power?" The key here is, is to use some basic trigonometric identities. What do I mean by that? We know that sin squared
X plus cosine squared X is equal to one, or if
we subtract sin squared from both sides, we know
that cosine squared X is equal to one, write
it this way, is equal to one minus sin squared X. What would happen if
cosine to the third power, that's cosine squared times cosine. What happens if we were to take that cosine squared? Let me just rewrite it. This is the same thing as cosine of X times cosine squared of X, DX. What if we were to take this thing right over here, let me
do that magenta color. What if we were to take
this right over here and replace it with this. I now what you're thinking. "Sal, what's that going to do for me? "This feels like I'm making this integral "even more convoluted." What I would tell you, I would say, "This might seem like it's
getting more complicated, "but as you explore and you play with it, "you'll see that this actually makes "the integral more solvable." Let's try it out. If we do that, this is
going to be equal to the indefinite integral,
cosine of X times one minus sin squared X, DX. What is this going to be equal to? This is going to be equal to, let me do this in that green color. This is going to be equal to the indefinite integral of cosine X. I'm just going to
distribute the cosine of X. Cosine of X minus, minus cosine of X, cosign of X sin squared of X, sin squared, sin squared X and then I can
close the parentheses, DX. This, of course, is going
to be equal to the integral of cosine of X, DX, and we know what that's going to
be, minus the integral. I'll switch to one color
now, of cosine of X, sin squared X, sin squared X, DX. Now, this is where it gets interesting. This part right over here
is pretty straight forward. The anti-derivative of
cosine of X is just sin of X. This right over here is
going to be sin of X. I'll worry about the plus C at
the end because both of these are going to have a
plus C, so might as well just put one big plus c at the end. That's sin of X and then what do we have going on over here? Well, you might recognize, I
have a function of sin of X. I'm taking sin of X and I'm squaring it and then I have sin of X's
derivative right over here. This fits the, I have some
derivative of a function and then I have another and then I have a, I guess you could say, a
function of that function. G of F of X. That's a sin that maybe U
substitution is in order, or we've seen the pattern,
we've seen this show multiple times already,
that you could just say, "Okay, if I have a function of a function "and I have that functions derivative, "then essentially I can just
take the anti-derivative "with respect to this function." This would be equal to, say, capital G is the anti-derivative of lower case G. Capital G of F of X plus C. Now, if what I said didn't
make sense, then we could do U substitution and go through it a little bit more step by step. Let's just do that because
we want things to make sense. That's the whole point of these videos. We could say U is equal
to sin of X and then DU is going to be equal to cosine of X, DX. This part and that part is going to be DU and then this is going to be U squared. This is going to be minus. We have the integral of U squared, DU. What is this going to be? This is going to be, we're
going to have negative U to the third power over three. Then, we know what U is. The U is equal to sin of X. We have our sin of X
here for the first part of the integral, for the first integral. We have the sin of X and then
this is going to be minus. Let me just write it this way. Minus 1/3 minus 1/3. Instead of U to the third,
we know U is sin of X. Sin of X to the third power. Then now, we can throw that plus C there. We're done. We've just evaluated
that indefinite integral. The key to it is to just
play around a little bit with trigonometric identities
so that you can get the integral to a point that you can use the reverse chain rule or
you can use U substitution, which is just really another way of expressing the reverse chain rule.