simpleetal
Member
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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 substitutionwould have made the first part a bit easier (no need for a second substitution). Rest looks fine though
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substitute
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
cosh and sinh?Using the substitutionshould work too, you will still need to use IBP twice to get the desired value.