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suju101

  • 2 years ago

integrate sin(tan^-1x)/(1+x^2). please help with the steps

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  1. imperialist
    • 2 years ago
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    Set u=arctan(x), so that du= 1/(1+x^2). Thus, using this u-substitution, the integral is equivalent to the integral of sin(u). Since the integral of sin(u) is -cos(u)+C, the final answer is just \[\sin (\tan^{-1} (x))+C\] Note also that sin(arctan(x)) is the same as \[x/\sqrt{x^2+1}\] So that function plus your constant of integration is also a correct solution.

  2. imperialist
    • 2 years ago
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    Sorry, it should be \[-\cos (\tan^{-1} (x))+C\]

  3. suju101
    • 2 years ago
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    thanx

  4. imperialist
    • 2 years ago
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    Also, that means that the alternate answer is instead \[1/\sqrt{x^2+1} + C\]

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