Indefinite Integral Involving Powers of Sines and Cosines

Key Strategy

• Odd Power Indicator: The strategy depends on whether sine or cosine has an odd power.
  • Odd Power of Cosine: Save one cosine factor and convert the rest to sines.
    • Let ( u = \sin x ).
    • Differential ( du = \cos x \ dx ).

Example

• Given: ( \sin^2 x \cdot \cos^5 x ).
  • Power of Cosine is Odd: Save one cosine factor.

Steps to Solve

1. Rewrite the Integral

   • ( \int \sin^2 x \cos^5 x \ dx )
   • Rewrite as ( \int \sin^2 x \cdot \cos^4 x \cdot \cos x \ dx )

2. Substitution

   • Let ( u = \sin x ), hence ( du = \cos x \ dx ).
   • Substitute in terms of ( u ):
     • ( \sin^2 x = u^2 )
     • ( \cos x dx = du )
   • Remaining ( \cos^4 x ) needs to be converted using the identity:
     • ( \cos^2 x = 1 - \sin^2 x )
     • ( \cos^4 x = (1 - \sin^2 x)^2 )

3. Convert and Simplify

   • Integral becomes ( \int u^2 (1 - u^2)^2 du )
   • Expand ( (1 - u^2)^2 = 1 - 2u^2 + u^4 )
   • Distribute ( u^2 ):
     • ( u^2(1 - 2u^2 + u^4) = u^2 - 2u^4 + u^6 )

4. Integrate Using Power Rule

   • ( \int (u^2 - 2u^4 + u^6) du )
   • Apply power rule:
     • ( \frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7} + C )

5. Substitute Back

   • Replace ( u ) with ( \sin x ):
     • ( \frac{1}{3} \sin^3 x - \frac{2}{5} \sin^5 x + \frac{1}{7} \sin^7 x + C )

Conclusion

• Key Concept: When cosine has an odd power, save a cosine factor, let ( u = \sin x ).

• Final antiderivative: ( \frac{1}{3} \sin^3 x - \frac{2}{5} \sin^5 x + \frac{1}{7} \sin^7 x + C ).

• Reminder: The method applies similarly with sine having odd power by saving one sine factor and converting cosines.