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Math Help - Integration with trig functions

  1. #1
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    Integration with trig functions

    Hi I'm stuck on a problem (I don't know if I am doing it correctly):

    \int \frac{1}{1 + x^2}.dx
    I have been given a hint to substitute x^2 with tan^2(u).

    When I do this it becomes \frac{1}{1 + tan^2(u)}.dx now using the identity of tan^2 = sec^2 - 1 i get \frac{1}{\sec^2(u)}.dx it becomes \cos^2(u).dx.

    I know that to make it so the integral is with respect to (u) I find the differential of x= tan^2(u) which when I rearrange becomes dx = sec^2(u).du therefore \int cos^2(u)\sec^2(u).du. Here is where I am confused, does the above just become \int 1 because sec and cos cancelling each other out? or is there something that I have missed out?

    Thanks for any help
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  2. #2
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    Quote Originally Posted by Beard View Post
    Hi I'm stuck on a problem (I don't know if I am doing it correctly):

    \int \frac{1}{1 + x^2}.dx
    I have been given a hint to substitute x^2 with tan^2(u).

    When I do this it becomes \frac{1}{1 + tan^2(u)}.dx now using the identity of tan^2 = sec^2 - 1 i get \frac{1}{\sec^2(u)}.dx it becomes \cos^2(u).dx.

    I know that to make it so the integral is with respect to (u) I find the differential of x= tan^2(u) which when I rearrange becomes dx = sec^2(u).du therefore \int cos^2(u)\sec^2(u).du. Here is where I am confused, does the above just become \int 1 Mr F says: No. It becomes {\color{red}\int 1 du = u + C} ....

    because sec and cos cancelling each other out? or is there something that I have missed out?

    Thanks for any help
    ..
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