Hi, Could someone walk me through this question: Evaluate the improper integral sinh2x/1+cosh^4x dx between 0 and infinity. So far I have changed sinh2x to 2sinhxcoshx and I think I will use u substitution and make u=coshx Also du=sinhx, so I have u.du/1+u^4...does that mean I integrate to get (u^2/2) / x+u^5/5 ? And then I get to 5u^2/x+u^5... Then for x i have cosh-1u? So overall it is 5u^2/cosh-1u+u^5

Question

experimentx · Answer

try changing $ \cosh^4 x  = (\cosh^2x)^2 = (\cosh 2x + 1)^2 $ use use substitution $ \cosh 2x = u $

anonymous · Answer

Thanks primeralph. The original equation was 1+cosh^3x as the denominator, does this change anything?

anonymous · Answer

1+cosh^4x sorry.

rajee_sam · Answer

I don't know maybe I am wrong..

$$put \space u = \cosh ^{2} x$$$$now \space du = -2\cosh x \sinh x$$ so now I have to integrate$$- \int\limits_{}^{} \frac{ du }{ 1 + u ^{2}} = - (\tanh ^{-1} u) + C = - (\tanh ^{-1}(\cosh ^{2} x) ) + C$$

??? Maybe I am not sure if such thing exists.....

rajee_sam · Answer

@Euler271

anonymous · Answer

When I have u=coshx I eventually get out (tan-1u^2)/2

anonymous · Answer

it seems right. and such a function can exist. cos(sin(tan(sec(pi)))) exists for example.

however, i just checked in on wolfram, and they show:

$$-(\tanh^{-1}(sech^{2}(x)) + C$$


and remember, since its imprper you're really taking:

$$\lim_{t->\infty} \int\limits_{0}^{t} simplification dx$$