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    • Thread Starter

    Can anyone help with this Q;

    \int \frac{1+sinx}{cosx}dx; u=sinx

    What do i do? Do i first get rid of the fraction, and say it's \int cosx^{-1}(1+sinx)dx Do I the times this out?

    So u=sinx, therefore \frac{du}{dx}=cosx. What does dx equal though? Is it dx=\frac{du}{cosx}? Is there a better way to put this?

    Can anyone help, I really don't know where to start?

    you end up with (1+u) / (cos^2 x) which is (1+u) / ( 1- sin ^2 x) which is (1+u) / (1 - u ^2) etc
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Updated: March 21, 2009

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