P5 problems

I have a few questions that i'm having trouble with. I'd be grateful if anyone could correct either myself or the book.

(i) Use the substitution u=e^x to find INT 1/(5coshx + 4sinhx) dx
The answer i get is (1/6)arctan(3e^x) + C The book gets (2/3)arctan(3e^x).

(ii) INT (x^2)/(1+x^2) dx with an upper limit rt3 and a lower limit of 1.
I get the answer 0.522 while the book gets the answer 0.470.

Thanks for any help.

Thanks. I'll post my working:

INT 1/(5coshx + 4sinhx) dx
= (1/2) INT 1/[5e^x + 5e^-x + 4e^x - 4e^-x]
= (1/2) INT 1/(9e^x + e^-x)
= (1/2) INT e^x/9e^2x + 1
= (1/18) INT e^x / e^2x + (1/9)
u = e^x, du/dx = e^x

= (1/18) INT 1/(u^2 + 1/9) du
= (1/18) [3arctan3u] = (1/6)arctan3e^x + C

You've bypassed the fact that sinh(x) = [exp(x) - exp(-x)]/2
and cosh(x) = [exp(x) + exp(-x)]/2.

There should be a 2 in the numerator.

Galois.

(ii)

Taking x = sinhu. u=arsinhx dx/du = coshu, dx = coshu du

= INT (sinh^2 u)/(cosh^2 u) . coshu . du with limits arsinhrt3 and arsinh1

= INT (coshu - sechu) du
= [sinhu - 2arctane^u] Limits arsinhrt3, arsinh1.

Just integrate by parts,

I = ∫ cost/(sint + cost) dt = ∫ (sint + cost - sint)/(sint + cost) dt = ∫ 1 - sint/(sint + cost) dt
I = x + ln(sint + cost) - ∫ cost/(sint + cost) dt
I = x + ln(sint + cost) - I
2I = x + ln(sint + cost)
I = ½(x + ln(sint + cost) | t=0, t=π/2
I = ½{(π/2 + ln(1 + 0)) - (0 + ln(0 + 1))}
I = ½{π/2}
I = π/4
=====

Edit: you have to manipulate it first. Integrating the expression, ∫ cost/(sint + cost) dt by parts, as it is, doesn't work.

Firstly, thanks for the help.
I'm sorry but i dont see where the ln(sint+cost) has come from, i didn't see a function and derivative, ie, (cost-sint)/(sint+cost) as u cancelled off part of it earlier to make 1.