Hi! Please help me to answer this integral, in detailed

Question

It is a fraction so it goes something like this

Integral of (e^2x)(1-e^4x)^-1/2 all over (arcsin e^2x)cos²(ln (arcsin e^2x)) (dx on the middle of the fraction)

Please provide me with a detailed step so I can study it. Thank you.

Answer 1

I = ∫(e^2x)(1 - e^4x)^-1/2/[(arcsine^2x)cos²(ln(arcsine^2x))]dx;

Let du substitution u = e^2x, then du = 2e^2xdx; then I = 1/2∫(1/√(1 - u²))arcsinucos²(ln(arcsinu))du =

then v = arcsinu; then dv = 1/√(1-u²0du and I = 1/2∫1/[vcos²(lnv)]dv = 1/2tan(lnv) + C = 1/2tan(ln(arcsinu)) + C = 1/2tan(ln(arcsine^2x)) + C

Answer 2

We are going to make two substitutions

The first one is t=arcsin(e^2x). We differentiate both sides (pay attention to use the chain rule correctly!) and we obtain dt=(1/sqrt(1-e^4x))*2*e^2x You can notice that this quantity is exactly (apart from the factor 2) the numerator of your given integral.

If you do this (and you manage the factor 2), the integral which you need to evaluate is (1/2)*integral of dt/(t*cos²(log(t)))

We now make another substitution, y=ln(t). If we differentiate both sides we have dy=1/t dt and therefore, the previous integral becomes (1/2)*integral dy/cos²(y) which can be immediately solved and it is (1/2)tan(y).

We now go back with the substitutions and we first write the result as a function of the variable y=ln(t) as

(1/2)tan(ln(t)) and then we write it as a function of x (remember that t=arcsin(e^2x)) as (1/2)tan(ln(arcsin(e^2x))) +C, where for completeness we add the constant of integration.

Best,

Davide