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Calculus/Calculus II integration by parts

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Expert: - 2/6/2011


Question
How should I compute this integral? What would be the answer?

S (x^5)(cos^3(x^6)(sin^2(x^6)) dx

Answer
The first thing to do is let u = x^6, so du = 6*x^5 dx.  This is the samme as (1/6)du = x^5 dx.
This gives S cos^3(u)sin^2(u)du.

When powers of sin() and cos() are involved, only one of them should have a power greater than 1.
Since cos^2(u) = 1 - sin^2(u), we can now substitute that in, giving
S cos(u)(1-sin^2(u))sin^2(u) du.

Multiplying the 2nd and 3rd terms together gives S cos(u)(sin^2(u)-sin^4(u)) du.
Now we only have cos() to the 1st.

For the last step, let v = sin(u), so dv = cos(u) du.
Since the integral has a cos(u) at the start and a du at the end, this leaves us with
S v^2 - v^4 dv.

That is v^3/3 - v^5/5.  Since v = sin(u), that is sin^3(u)/3 - sin^5(u)/5.
Now since u = x^6, that becomes sin^3(x^6)/3 0 sin^5(x^6)/5,
and then being an integral with no limits, put a +C at the end.

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