Solve for the Anti-Derivative of cosx dx / sin^3x: Expert Answer and Explanation

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Homework Help Overview

The discussion revolves around finding the anti-derivative of the function cos(x) / sin^3(x), a topic within integral calculus.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants share their attempts at solving the integral, with one confirming their result aligns with a computational tool. Another suggests verifying the solution by taking the derivative of the result.

Discussion Status

The conversation includes various approaches to the problem, with one participant providing a substitution method. There is acknowledgment of the correctness of the solution, but no explicit consensus is reached on the overall discussion.

Contextual Notes

Participants reference the use of computational tools and suggest checking work through differentiation, indicating a focus on verification in the learning process.

laker_gurl3
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was just wondering if this was the right answer..
take the anti-derivative of:

cosx dx / sin^3x

i got
-1/2(sinx)^-2 + C
is that right?
 
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Thats what maple says, good job.
 
Just a tip, if your not sure, take the derivative and check if you get the original function.
 
Maple?? Maple?? ! Let u= sin x. Then du= cos du and the integrand becomes
[tex]\frac{cos x}{sin^3 x}dx= \frac{1}{u^3}du= u^{-3}du[/tex]

The anti-derivative of that is [tex]\frac{1}{-3+1}u^{-3+1}+C= \frac{1}{2}u^{-2}+C= \frac{1}{2}\frac{1}{sin^2 t}+C[/tex]

Good job, laker_gurl3
 
Nothing wrong with being lazy!
 

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