C4 Integration help

how would you integrate sin^2tcost??
I was looking on exam solutions and he seemed to use the method where you easily can spot the pattern, i find this method hard and wonder if there is another way round it?

how would you integrate sin^2tcost??
I was looking on exam solutions and he seemed to use the method where you easily can spot the pattern, i find this method hard and wonder if there is another way round it?

The one that jumps out at me is integration by parts but it'd be a real pig.

Practice questions and you should start spotting them.

You need to use a substitution u, so that $\frac{du}{dt}$ is equal to whatever is multiplied by the function of u.

In this case, the substitution is $u = \sin t$ because when differentiated, this become $\cos t$, which cancels with the $\cos t$ in the integrand.

The one that jumps out at me is integration by parts but it'd be a real pig.

Practice questions and you should start spotting them.

how did you know that u=sint.. this would have taken me a while to figure out.. guess practice is the key???

how did you know that u=sint.. this would have taken me a while to figure out.. guess practice is the key???

As I said ... That understanding is identical to the understanding needed to use inverse chain rule

Just because $\frac{d(\sin t)}{dt} = \cos t$.
You need to find a substitution that cancels the other term (that cannot be put in terms of the substitution).


After practice, you will be able to use inverse chain rule, which is quicker and based on intuition.

okay thank you both!! I will practice more of these

s(sin^2tcost) dx
u= (sint)^2
....

how would you integrate sin^2tcost??
I was looking on exam solutions and he seemed to use the method where you easily can spot the pattern, i find this method hard and wonder if there is another way round it?