There are a few possible scenarios for the integrand in the form sinαxcosβx:
Case 1: α is odd, β is even
Evaluate ∫sin3xcos2xdx.
Rewrite it as ∫sinxsin2xcos2xdx=∫sinx(1−cos2x)cos2xdx.(1) Let u=cosx, then dxdu=−sinx and (1) can be rewritten as −∫(1−u2)u2du. This integral is very easy to compute now; you just have to expand the terms and use the power rule for antiderivatives. But don't forget what was u! □
Case 2: α is even, β is odd
Same idea as "α is odd, β is even."
Case 3: both α and β even
In this scenario, there are two different things you could do. You could utilize the following identities:
- cos2x=21+cos2x
- sin2x=21−cos2x.
Or, you could rewrite the integrand only in terms of a single trigonometric function.
Evaluate ∫sin2xcos2xdx.
Using the aforementioned identities, we can rewrite the given expression as
∫(21−cos2x)(21+cos2x)dx=41∫(1−cos22x)dx=41∫sin22xdx.(1)
Let u=2x, thendxdu=2 and (1) can be rewritten as
81∫sin2udu.(2)
Remember? We got an identity for this! So (2) can be rewritten as
81∫21−cos2udu=16u−32sin2u+C,
where C is the constant of integration.
But what was u? Substituting u=2x, the answer is
8x−32sin4x+C. □
Something to consider: What's another way you could solve that integral? Since the powers are relatively nice, you could rewrite everything in terms of cosine or sine, and use one of the reduction formulas from above.
Evaluate ∫sin2xcos4xdx.
It is possible to use the aforementioned problems to solve this problem, but I'm going to use the reduction formulas approach just to contrast the methods. For this type of problem, I actually prefer the reduction formulas:
∫sin2xcos4xdx=∫(1−cos2x)cos4xdx=∫cos4xdx−∫cos6xdx.(1)
Since ∫cos6dx, which is the integral with the highest exponent in (1), can be reduced to
∫cos6dx=6cos5xsinx+65∫cos4xdx,(2)
substituting (2) into (1) gives
∫sin2xcos4xdx=61∫cos4dx−6cos5xsinx.(3)
Use the reduction formulas again, then ∫cos4xdx in (3) becomes
∫cos4dx=4cos3xsinx+43∫cos2xdx.(4)
We know that cos2x=21+cos2x and we also know how to integrate that, so I'm going to skip that step and find (4) to be equivalent to ∫cos4dx=4cos3xsinx+43(2x+4sin2x).(5)
Substituting (5) into (3) gives ∫sin2xcos4xdx=24cos3xsinx+16x+32sin2x−6cos5xsinx+C,
where C is the constant of integration. □