Integration using inverse trig indentities?

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Homework Help Overview

The discussion revolves around two integration problems involving inverse trigonometric identities. The first integral involves the function sin(x) in the numerator and a composite function in the denominator, while the second integral features a quadratic expression in the denominator.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various approaches to tackle the integrals, including substitution methods and completing the square. There is uncertainty regarding the effectiveness of the initial attempts, particularly with the first integral's numerator.

Discussion Status

Some participants have suggested specific substitutions for the first integral, while others have pointed out that a second substitution may not be necessary. For the second integral, guidance has been offered to complete the square in the denominator, indicating a potential direction for further exploration.

Contextual Notes

Participants are navigating through the complexities of the integrals, with some expressing difficulties in their initial attempts and questioning the assumptions underlying their approaches.

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Homework Statement


1.\int{\frac{sinx}{1+cos^{2}x}} \, dx
2.\int{\frac{1}{13-4x+x^2}} \, dx

Homework Equations


Inverse trig identities.

The Attempt at a Solution


For the first one, I'm not too sure about what to do with the sinx on the numerator and i have tried u-substitution to no avail (different answer). For the second one, I tried factorising but only had solutions for x\inℂ.
 
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You can maybe try to start with the substitution u = cos(x) and reforming the integral. A second substitution will be necessary after the first substitution.
 
eple said:
You can maybe try to start with the substitution u = cos(x) and reforming the integral. A second substitution will be necessary after the first substitution.
Second substitution is not necessary. It can be done in that first one.
 
For the second integral start by completing the square in the denominator.
 

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