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TomLikesPhysics
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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.
 2 years ago
 2 years ago
TomLikesPhysics Group Title
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.
 2 years ago
 2 years ago

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NotSObright Group TitleBest ResponseYou've already chosen the best response.1
Replace cos x by sqrt(1t^2)
 2 years ago

NotSObright Group TitleBest ResponseYou've already chosen the best response.1
Put sin x = t
 2 years ago

NotSObright Group TitleBest ResponseYou've already chosen the best response.1
But going by this method is possible and feasible but not smart
 2 years ago

NotSObright Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{}^{}\frac{t^2}{\sqrt{1t^2}}dt\]
 2 years ago

NotSObright Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{}^{}\sqrt{1t^2}\frac{1}{\sqrt{1t^2}}\]
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.1
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 deadend.
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}t ^{2}/\sqrt{1t ^{2}}=\int\limits_{}^{}(1t ^{2}1)/\sqrt{1t ^{2}}\] =\[\int\limits_{}^{}(1t ^{2})/\sqrt{1t ^{2}}1/\sqrt{1t ^{2}}\]. That's how he got there
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.1
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 0.50.5*cos2x.
 2 years ago
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