well, this can be written as [ ] [ ] sin2x/((a^2-b^2)sin 2x + b^2)) dx now put sin2x = t u will get sin2x dx = dt so it becomes [ ][ ] dt/((a^2-b^2)t^2 + b^2) = 1/(a^2-b^2) [ ][ ] dt/(t 2 + b/ [ ] (a^2-b^2)) =1/b [ ](a 2-b 2) tan -1(t [ ](a 2-b 2)/b) + c now put t= sin2x in this and get the answer Please rate !
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Moderation Team
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