simpleetal
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- Apr 6, 2015
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- 2016
Last edited:
You have left out all the important working required to get that result ... ie. using the tan(A-B) result
I would like to agree^ I think using the direct substitution would have made the first part a bit easier (no need for a second substitution). Rest looks fine though
substitute , then carry out IBP twice. i got
Won't the first IBP be u = 1/(tantheta + sectheta)^n and dv = sec^2 thetasubstitute , then carry out IBP twice
yes thats same as i didWon't the first IBP be u = 1/(tantheta + sectheta)^n and dv = sec^2 theta
but the borders will be from pi/4 to pi/2 and when you get uv, you have to sub theta = pi/2 and that gives infinityyes thats same as i did
no, at pi/2, you still get finite, actually zero ( by multiplying cos^n both numerator and denominator )but the borders will be from pi/4 to pi/2 and when you get uv, you have to sub theta = pi/2 and that gives infinity
Using the substitution should work too, you will still need to use IBP twice to get the desired value.
cosh and sinh?Using the substitution should work too, you will still need to use IBP twice to get the desired value.