Vtyjkyuk

I would need to solve this integral analytically using also a software.
Can you help me?

$$int_-2^x(x-varepsilon)frac(varepsilon+2)^1.5(2-varepsilon)^1.52.5^-2 dvarepsilon - x = a$$

Let $I(x)=int_-2^x(x-varepsilon)frac(varepsilon+2)^1.5(2-varepsilon)^1.52.5^-2 dvarepsilon$

Rewrite the integrand as

$$I(x)=2.5^2int_-2^x (x-epsilon)(4-epsilon^2)^1.5 depsilon$$

Then, by expanding $(x-epsilon)$ into two separate integrals and manipulating, we have

$$I(x)=2.5^2xint_-2^x (4-epsilon^2)^1.5 depsilon-2.5^2int_-2^x epsilon(4-epsilon^2)^1.5 depsilon$$

The first integral can be solved by substituting $epsilon=2sinphi, depsilon=2cosphi dphi$, and using the fact that $cos^4 phi = frac12cos2phi+frac18cos4phi+frac38$ and $1-sin^2phi = cos^2phi$

The second integral can be solved easily by substituting $u=4-epsilon^2,du=-2epsilon depsilon$

Note: Be careful in changing and substituting the limits when performing a change in variables or substituting after you have found the antiderivative.

Hint:

With $epsilon=2costheta$,

$$int(x-epsilon)(4-epsilon^2)^3/2depsilon=-16int(x-2costheta)sin^3thetasintheta,dtheta.$$

The second term is immediate, $$-frac325sin^5theta=-frac325(4-epsilon^2)^5/2.$$

The first can be addressed with

$$sin^4theta=left(frac1-cos2theta2right)^2=frac14-cos2theta+frac14cos^22theta=frac14-cos2theta+fraccos4theta+18.$$

Integrate and convert back to a function of $epsilon$.