Help with trigonometric integral

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SUMMARY

The integral \(\int_{\frac{5\pi}{6}}^{\pi}\frac{\cos^{4}x}{\sqrt{1-\sin{x}}}dx\) can be solved using a substitution method. The suggested substitution is \(u=1-\sin x\), which simplifies the integral significantly. The equation \(\cos^{2}x=\frac{1+\cos{2x}}{2}\) is also useful for expanding \(\cos^{4}x\) to facilitate integration. This approach leads to a more manageable integral that can be solved using standard techniques.

PREREQUISITES
  • Understanding of trigonometric identities, specifically \(\cos^{2}x\) and \(\sin x\)
  • Familiarity with integration techniques, particularly substitution methods
  • Knowledge of definite integrals and their evaluation
  • Basic proficiency in calculus, particularly in handling trigonometric functions
NEXT STEPS
  • Practice solving integrals involving trigonometric functions using substitution
  • Explore advanced integration techniques such as integration by parts and trigonometric substitution
  • Learn about the properties of definite integrals and their applications
  • Study the derivation and application of trigonometric identities in integration
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for effective methods to teach trigonometric integrals.

miglo
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Homework Statement


[tex]\int_{\frac{5\pi}{6}}^{\pi}\frac{\cos^{4}x}{\sqrt{1-\sin{x}}}dx[/tex]

Homework Equations


[tex]\cos^{2}x=\frac{1+\cos{2x}}{2}[/tex]


The Attempt at a Solution


i used the above equation, then expanded it all out and multiplied by the denominator and hoped i would then be able to do a simple substitution that would give me an antiderivative after integrating but that hasn't been working for me
any ideas?
 
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miglo said:

Homework Statement


[tex]\int_{\frac{5\pi}{6}}^{\pi}\frac{\cos^{4}x}{\sqrt{1-\sin{x}}}dx[/tex]

Homework Equations


[tex]\cos^{2}x=\frac{1+\cos{2x}}{2}[/tex]


The Attempt at a Solution


i used the above equation, then expanded it all out and multiplied by the denominator and hoped i would then be able to do a simple substitution that would give me an antiderivative after integrating but that hasn't been working for me
any ideas?
A relatively straightforward method involves a series of substitutions. I would start with [itex]u=1-\sin x[/itex]
 

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