How Do You Solve the Integral of sin^3(x)cos^6(x)dx?

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SUMMARY

The integral of sin^3(x)cos^6(x)dx is solved using the substitution method, where u = cos(x) and du = -sin(x)dx. The integral simplifies to -∫(u^63 - u^65)du, resulting in the final expression of (-cos^64(x)/64) - (cos^66(x)/66) + c. Verification through differentiation confirms the correctness of the solution, ensuring that the original integrand is recovered.

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  • Familiarity with trigonometric identities and their derivatives
  • Knowledge of integration techniques, particularly polynomial integration
  • Ability to differentiate functions to verify integration results
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  • Study advanced integration techniques, including integration by parts
  • Explore trigonometric integrals and their simplifications
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Homework Statement


∫sin^3xcos^63xdx


Homework Equations





The Attempt at a Solution


∫sin^3xcos^63xdxsin^3xcos^63xdx

let u=cosx
du=-sinxdx
-du=sinxdx

=∫sin^2xcos^63xsinxdx
=∫(1-cos^2x)cos^63x-du
=-∫(1-u^2)u^63du
=-∫(u^63-u^65)du
=-u^64/64-u^66/66+c
=(-cos^64x/64)-(cos^66x/66)+c

^ I'm thinking this is not the correct answer?
 
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Try differentiating it and see if you recover the integrand.
 

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