integral of sin(x)*cos^3(x) vs integral of sin^2(x)*cos^3(x), calculus 2 tutorial | Summary and Q&A

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February 13, 2015
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integral of sin(x)*cos^3(x) vs integral of sin^2(x)*cos^3(x), calculus 2 tutorial

TL;DR

This content explains the process of integrating sine and cosine functions using different methods.

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Questions & Answers

Q: How can the first integral, involving sine and cosine functions, be solved easily?

The first integral can be solved easily by using a substitution method. By letting U equal to cosine X, the integral simplifies, and after applying the power rule, the solution is obtained.

Q: Why is the second integral, involving sine squared and cosine cubed, trickier to solve?

The second integral is trickier because direct cancellation of the sine squared and cosine cubed terms is not possible. Trigonometric identities are required to break down the expression and then substitution can be applied to simplify the integral.

Q: How does the use of trigonometric identities help in solving the second integral?

Trigonometric identities, specifically the identity that states sine squared x is equal to 1 minus cosine squared x, can be used to simplify the second integral. By breaking down the expression and applying the identity, the integral can be transformed into a more manageable form.

Q: How is the second integral solved after applying trigonometric identities and substitution?

After applying the trigonometric identity and substitution (U = sine X), the integral becomes a straightforward integration of U squared minus U to the fourth power. By using the power rule and substituting U back to sine X, the final solution is obtained.

Summary & Key Takeaways

  • The first integral involves sine of x multiplied by cosine to the third power of x, which can be easily solved using a substitution method.

  • The second integral involves sine squared x multiplied by cosine to the third power of x, which requires the use of trigonometric identities and substitution.

  • By applying the power rule and appropriate substitutions, the integrals can be simplified to obtain the final solutions.

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