Question

If I want to integrate (sinx)^2 using the method of substitution than what are my function g and g`? x would be pointless because than I would just replace the x with a t but sinx also does not seem to work because there is no cosx.

Answer 1

Replace cos x by sqrt(1-t^2)

Answer 2

Put sin x = t

Answer 3

But going by this method is possible and feasible but not smart

Answer 4

\[\int\limits_{}^{}\frac{t^2}{\sqrt{1-t^2}}dt\]

Answer 5

\[-\int\limits_{}^{}\sqrt{1-t^2}-\frac{1}{\sqrt{1-t^2}}\]

Answer 6

How do you get from one term (the t^2 over sth.) to the two separate terms? At first I followed my instincts and got to the t^2 over such and such term but I thought this is a dead-end.

Answer 7

\[\int\limits_{}^{}t ^{2}/\sqrt{1-t ^{2}}=\int\limits_{}^{}-(1-t ^{2}-1)/\sqrt{1-t ^{2}}\]
=\[-\int\limits_{}^{}(1-t ^{2})/\sqrt{1-t ^{2}}-1/\sqrt{1-t ^{2}}\].
That's how he got there

Answer 8

Ok, that would have never come to my mind. Thank you for your explanation. I just figured out that it is quite easy to do this integral if I write instead of (sinx)^2 just 0.5-0.5*cos2x.