Trigonometric Integral: Solving (5pi/6 to pi) (cosx)^4 / sqrt(1-sinx) dx

[whatlifeforme]

Homework Statement

solve the integral.

Homework Equations

integral (5pi/6 to pi) (cosx)^4 / sqrt(1-sinx) dx

The Attempt at a Solution

Following the solution manual:

integral (5pi/6 to pi) ( (cosx)^4 / sqrt(1-sinx)) * sqrt(1+sinx)/sqrt(1+sinx)

however i am not sure how to do the algebra here.

it should convert to integral (5pi/6 to pi) ( (cosx)^4 / sqrt(1-sin^2x)) * sqrt(1+sinx) dx

 

[LCKurtz]

Homework Statement

solve the integral.

Homework Equations

integral (5pi/6 to pi) (cosx)^4 / sqrt(1-sinx) dx

The Attempt at a Solution

Following the solution manual:

integral (5pi/6 to pi) ( (cosx)^4 / sqrt(1-sinx)) * sqrt(1+sinx)/sqrt(1+sinx)

however i am not sure how to do the algebra here.

it should convert to integral (5pi/6 to pi) ( (cosx)^4 / sqrt(1-sin^2x)) * sqrt(1+sinx) dx

When you say "it should convert" do you mean you don't see how do get that last step or you don't know what to do next? You do know that $1-\sin^2 x = \cos^2 x$, right? You could change all the even powers of $\cos x$ to sines and do a u substitution.

 

[whatlifeforme]

solved.

how about..

interal (sin2x)^4 / (sqrt(1-cos2x) dx

 

[LCKurtz]

solved.

how about..

interal (sin2x)^4 / (sqrt(1-cos2x) dx

Those problems are so similar that whatever you did for the first one (which you haven't shown us) should work on the second one.

 

[whatlifeforme]

i can't remember what i did and i lost the scrap piece of paper. i think i multiplied by the conjugate but I am stuck.

 

[SammyS]

i can't remember what i did and i lost the scrap piece of paper. i think i multiplied by the conjugate but I am stuck.

That should work.

What are you stuck with?

 

[whatlifeforme]

after multiplying by the conjugate and simplifying i get: integral (sin2x)^3 * sqrt(1+cos2x) dx

 

[SammyS]

after multiplying by the conjugate and simplifying i get: integral (sin2x)^3 * sqrt(1+cos2x) dx

OK. So you have

\displaystyle \int (\sin(2x))^3\sqrt{1+\cos(2x)\,}\,dx\ .​

What did LCKurtz have you do in post #2 ?

...
You do know that $1-\sin^2 x = \cos^2 x$, right? You could change all the even powers of $\cos x$ to sines and do a u substitution.

This time all the arguments are 2x, rather than x, but so what?

 

[whatlifeforme]

there are no even power cosines.

 

[iRaid]

Got it.
Take u=2x and then you have to use the fact that sin^3x=sin^2x*sinx...

 

[SammyS]

there are no even power cosines.

There certainly is an even power of cosine.

cos³(2x)=cos²(2x)∙cos(2x).

There's a cos² for you.

 

[whatlifeforme]

Got it.
Take u=2x and then you have to use the fact that sin^3x=sin^2x*sinx...

what about the square root with 1+cos2x, though?

There certainly is an even power of cosine.

cos³(2x)=cos²(2x)∙cos(2x).

There's a cos² for you.

the problem is (sin2x)^3 not (cos2x)^3.

 

[whatlifeforme]

in my solutions manual, in has the following as part of the solution. how is this correct algebra?

(-2/3) + (2/3)(3/2)^(3/2) --> = sqrt(3/2) - 2/3

what happened to the 3 in the exponent (3/2)??

 

[Karnage1993]

(2/3) is the same as 1/(3/2) = (3/2)^-1.

 

[SammyS]

what about the square root with 1+cos2x, though?

the problem is (sin2x)^3 not (cos2x)^3.

My mistake !

(sin2x)^3 = (sin2x)²(sin2x)

     = (1-cos²(2x))sin(2x)

That all suggests to me the substitution, u = cos(2x) .

Added in Edit:

Perhaps better yet is u = 1 + cos(2x). Do you see why?

 

[LCKurtz]

@whatlifeforme: What I don't get about this thread is that in post #3 you claimed you have solved your original problem. Then you posted an almost identical problem just changing the sines to cosines and the x's to 2x's. So why are you going on and on with questions about every little step when you have already solved an essentially identical problem? What's going on here??

 

[iRaid]

what about the square root with 1+cos2x, though?



the problem is (sin2x)^3 not (cos2x)^3.

Then you do another u substitution and then an integration by parts from what I remember doing.