- #1
Adyssa
- 203
- 3
During maths class last semester this integral came up in the course of discussion and my lecturer gave a quick outline of how to solve it but I didn't grasp it at the time and we moved on. I'd like to know how to do it though! The integral is:
[itex]\int sin^{2}(x)[/itex]
The next step was:
[itex]\int sin(x).(-cos(x)')[/itex]
where [itex]-cos(x)'[/itex] denotes the derivative of [itex]-cos(x)[/itex] and the implication was that [itex]sin^{2}(x)[/itex] is the result of applying the chain rule to whatever compound function it is the derivative of. I can't remember the result but further steps were skipped and he jumped straight to the answer at this point.
I'm not sure how to proceed here, because the chain rule implies that [itex]\frac{dy}{dx} f(g(x)) = f'(g(x) . g'(x)[/itex], and in my example [itex]g(x) = x[/itex] and [itex]g'(x) = 1[/itex] which doesn't help me end up with [itex]sin(x)[/itex] as required.
Could anyone poke me in the right direction to solve this? =)
[itex]\int sin^{2}(x)[/itex]
The next step was:
[itex]\int sin(x).(-cos(x)')[/itex]
where [itex]-cos(x)'[/itex] denotes the derivative of [itex]-cos(x)[/itex] and the implication was that [itex]sin^{2}(x)[/itex] is the result of applying the chain rule to whatever compound function it is the derivative of. I can't remember the result but further steps were skipped and he jumped straight to the answer at this point.
I'm not sure how to proceed here, because the chain rule implies that [itex]\frac{dy}{dx} f(g(x)) = f'(g(x) . g'(x)[/itex], and in my example [itex]g(x) = x[/itex] and [itex]g'(x) = 1[/itex] which doesn't help me end up with [itex]sin(x)[/itex] as required.
Could anyone poke me in the right direction to solve this? =)