Isn't (1-sin²x)=cos²x?
You will end up with an expression containing cos(x) and ONE sin(x).
Try u=cos(x), du=d(cos(x))=-sin(x)dx
Do you see the light?
First make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals:
∫ sinx cos³x e^(1-sin²x) dx
I was going to substitute u= 1 - sin²x, but then i got du = ½ sinxcosx - ½ xdx, so im stuck.
Thank you so much!!
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1 answer