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Integral with trig-substitution

So I'm pretty much stuck with an integral problem. It was on the form (1 + (2t)^2)^0,5, using trig substitution I got to the integral 1/2(sec x)^3. Continuing with integration by parts, I finally got the solution (INT)1/2(sec x)^3 = 1/4(sec x)(tan x) + 1/4 ln |sec x + tan x| which should be correct. However, when I try to substitute back, using t = 1/2 tan x, my answer does not equal the answer in my book. And I can't seem to figure out what I did wrong... I've uploaded a picture of the last part of the problem, with the answer in my book below my answer.

Any help would be greatly appreciated!

Re: Integral with trig-substitution

Hello, gralla55!

Your book's answer is correct, but it's rather clumsy.

I would avoid those silly fractions . . .

And: .

Substitute: .

. . . . . . . . . .[1]

Back-substitute: .

is in a right triangle with: .

Hence: .

And: .

Substitute into [1]: .

. . . . . . . . . . . . .

Re: Integral with trig-substitution

Hi and thank you for very much for your help! I see you got the exact same answer I got, except I factored in the (1/2) under the first square root sign. What I can't seem to understand, is how the book gets from ln(2t + SQRT(1 + 4t^2)) to ln(t + SQRT((1/4 + t^2)). How are these two things the same??

Thanks again!

Re: Integral with trig-substitution

they're not "the same" ...

... but they do differ by a constant.

Re: Integral with trig-substitution

How did I not think of that.... Thank you very much! Now I can finally go to sleep, haha.