You need to evaluate ∫ sin²(x) cos³(x) dx. Which strategy is correct?
AApply power-reduction to both factors since sin has an even exponent
BSave cos(x) as the du factor, convert cos²(x) = 1 − sin²(x), then let u = sin(x)
CIntegrate by parts with u = sin²(x) and dv = cos³(x) dx
DApply the double-angle identity sin(2x) = 2 sin(x) cos(x) to the entire product
The exponent of cos is 3 (odd), so the odd-power strategy applies regardless of the sin exponent. Save one cos(x) for du, rewrite cos²(x) = 1 − sin²(x), and substitute u = sin(x). Option A is the classic misconception — students fixate on sin being even, but it is cos's odd exponent that determines the strategy. Power-reduction is only needed when both exponents are even.
Question 2 Multiple Choice
You encounter ∫ sin⁴(x) cos²(x) dx. Both exponents are even. The correct approach is:
ALet u = cos(x) and save sin(x) for du — the odd-exponent trick works on either factor
BApply power-reduction: sin²(x) = (1 − cos 2x)/2 and cos²(x) = (1 + cos 2x)/2, then expand
CUse integration by parts repeatedly, since no substitution applies when both exponents are even
DThe integral cannot be evaluated in closed form
When both exponents are even, no single factor can be 'saved' for a substitution du because saving one would still leave an odd power of the other. Power-reduction (half-angle) identities are the correct tool: they lower the powers while introducing double-angle terms that are eventually integrable. Integration by parts (option C) is a last resort here and far less efficient.
Question 3 True / False
In ∫ sinᵐ(x) cosⁿ(x) dx with m odd, the correct procedure saves one sin(x) factor as du and uses sin²(x) = 1 − cos²(x) to convert the remaining sin factors, then substitutes u = cos(x).
TTrue
FFalse
Answer: True
Exactly right. With m odd, write sinᵐ(x) = sinᵐ⁻¹(x) · sin(x). The sinᵐ⁻¹(x) piece has even exponent m−1, so it can be converted to a polynomial in cos²(x) using sin²(x) = 1 − cos²(x). The leftover sin(x) pairs with dx to form du = −sin(x) dx (with u = cos(x)), turning the integral into a polynomial in u — straightforward to integrate.
Question 4 True / False
If both exponents in ∫ sinᵐ(x) cosⁿ(x) dx are even, u-substitution with u = cos(x) can still reduce the integral to a manageable form.
TTrue
FFalse
Answer: False
U-substitution with u = cos(x) requires a sin(x) factor for du. With both exponents even, saving one sin(x) leaves sinᵐ⁻¹(x) — an odd power — which cannot be converted to a polynomial in cos²(x) using sin²(x) = 1 − cos²(x) without leaving a remaining sin(x). The correct tool for the both-even case is the power-reduction (half-angle) identities, which lower the powers algebraically without needing a du factor.
Question 5 Short Answer
Why does the odd-exponent strategy work for ∫ sinᵐ(x) cosⁿ(x) dx when one exponent is odd, but not when both are even?
Think about your answer, then reveal below.
Model answer: When one exponent is odd, you can split off one factor (e.g., sin(x)) to serve as du in a substitution. The remaining even-power factor converts cleanly to a polynomial in the other function via the Pythagorean identity. When both exponents are even, any attempt to save a factor leaves an odd power that can't be converted without creating another leftover factor — the substitution chain never terminates. Power-reduction identities bypass this by algebraically lowering powers without needing a du factor.
The substitution strategy works precisely because an odd exponent provides the 'extra' factor needed for du. Even exponents don't provide this, so the algebraic approach (power-reduction) becomes necessary. Recognizing which case applies — and why — is the central skill of trigonometric integration.