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

In summary: I think this is what I did in my lost work. In summary, the conversation is discussing how to solve an integral involving a trigonometric function. The student follows the solution manual, but is unsure of how to do the algebra. Another problem is presented and the student is advised to use a u substitution and the fact that sin^3x = sin^2x*sinx. The student then asks about the square root in the solution manual and is reminded of the rule (a/b)^-1 = b/a. The conversation ends with the student still unsure about the solution.
  • #1
whatlifeforme
219
0

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
 
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  • #2
whatlifeforme said:

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.
 
  • #3
solved.

how about..

interal (sin2x)^4 / (sqrt(1-cos2x) dx
 
  • #4
whatlifeforme said:
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.
 
  • #5
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.
 
  • #6
whatlifeforme said:
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?
 
  • #7
after multiplying by the conjugate and simplifying i get: integral (sin2x)^3 * sqrt(1+cos2x) dx
 
  • #8
whatlifeforme said:
after multiplying by the conjugate and simplifying i get: integral (sin2x)^3 * sqrt(1+cos2x) dx
OK. So you have
[itex]\displaystyle \int (\sin(2x))^3\sqrt{1+\cos(2x)\,}\,dx\ .[/itex]​

What did LCKurtz have you do in post #2 ?
LCKurtz said:
...
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?
 
  • #9
there are no even power cosines.
 
  • #10
Got it.
Take u=2x and then you have to use the fact that sin^3x=sin^2x*sinx...
 
Last edited:
  • #11
whatlifeforme said:
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.
 
  • #12
iRaid said:
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?

SammyS said:
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.
 
  • #13
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)??
 
  • #14
(2/3) is the same as 1/(3/2) = (3/2)^-1.
 
  • #15
whatlifeforme said:
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?
 
  • #16
@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??
 
  • #17
whatlifeforme said:
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.
 

1. What is a trigonometric integral?

A trigonometric integral is an integral that involves trigonometric functions, such as sine, cosine, and tangent.

2. What is the purpose of solving (5pi/6 to pi) (cosx)^4 / sqrt(1-sinx) dx?

The purpose of solving this trigonometric integral is to find the exact value of the area under the curve of the function within the given limits of integration.

3. What are the steps for solving this trigonometric integral?

The general steps for solving a trigonometric integral are: 1) Simplify the integrand using trigonometric identities, 2) Apply the appropriate substitution, 3) Evaluate the integral, and 4) Plug in the limits of integration and solve for the final answer. For this specific integral, the steps may vary depending on the chosen substitution.

4. Can this trigonometric integral be solved using integration techniques other than substitution?

Yes, there are other integration techniques that can be used to solve this integral, such as integration by parts or partial fractions. However, substitution may be the most efficient method for this specific integral.

5. What are some real-world applications of trigonometric integrals?

Trigonometric integrals have many real-world applications, including calculating the area under a curve in physics and engineering problems, finding the center of mass in geometric shapes, and solving differential equations that model real-world phenomena.

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