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TomLikesPhysics

  • 3 years ago

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.

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  1. NotSObright
    • 3 years ago
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    Replace cos x by sqrt(1-t^2)

  2. NotSObright
    • 3 years ago
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    Put sin x = t

  3. NotSObright
    • 3 years ago
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    But going by this method is possible and feasible but not smart

  4. NotSObright
    • 3 years ago
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    \[\int\limits_{}^{}\frac{t^2}{\sqrt{1-t^2}}dt\]

  5. NotSObright
    • 3 years ago
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    \[-\int\limits_{}^{}\sqrt{1-t^2}-\frac{1}{\sqrt{1-t^2}}\]

  6. TomLikesPhysics
    • 3 years ago
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    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.

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

  8. TomLikesPhysics
    • 3 years ago
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    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.

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