integration

âˆ«sin^-√x - cos^-√x / sin^-√x + cos^-√x dx = ?

5 Answers

it can also be written as,

-âˆ« tan(Ï€/4-√x)dx

now put √x=t

& apply integration be parts...............

sin-âˆšx + cos-âˆšx = pi/2

so the integration boils down to..

âˆ«pi/2 - 2cos-âˆšx / pi/2 dx = âˆ«dx - âˆ«4cos-âˆšx / pi dx

now simple by parts..

arey woh sab inverse hi hai naa.. ??

same question,

i have not taken it as inverse.

ok..

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