anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Replace cos x by sqrt(1-t^2)
anonymous
  • anonymous
Put sin x = t
anonymous
  • anonymous
But going by this method is possible and feasible but not smart

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.