Solve definite integral - solve analytically

Question

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

$$\int_{-2}^{x}(x-\varepsilon)\frac{(\varepsilon+2)^{1.5}(2-\varepsilon)^{1.5}}{2.5^{-2}} d\varepsilon - x = a$$

Answer 1

Let $I(x)=\int_{-2}^{x}(x-\varepsilon)\frac{(\varepsilon+2)^{1.5}(2-\varepsilon)^{1.5}}{2.5^{-2}} d\varepsilon$

Rewrite the integrand as

$$I(x)=2.5^2\int_{-2}^x (x-\epsilon)(4-\epsilon^2)^{1.5} d\epsilon$$

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

$$I(x)=2.5^2x\int_{-2}^x (4-\epsilon^2)^{1.5} d\epsilon-2.5^2\int_{-2}^x \epsilon(4-\epsilon^2)^{1.5} d\epsilon$$

The first integral can be solved by substituting $\epsilon=2\sin\phi, d\epsilon=2\cos\phi d\phi$, and using the fact that $\cos^4 \phi = \frac{1}{2}\cos2\phi+\frac{1}{8}\cos4\phi+\frac{3}{8}$ and $1-\sin^2\phi = \cos^2\phi$

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

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

Answer 2

Hint:

With $\epsilon=2\cos\theta$,

$$\int(x-\epsilon)(4-\epsilon^2)^{3/2}d\epsilon=-16\int(x-2\cos\theta)\sin^3\theta\sin\theta\,d\theta.$$

The second term is immediate, $$-\frac{32}{5}\sin^5\theta=-\frac{32}{5}(4-\epsilon^2)^{5/2}.$$

The first can be addressed with

$$\sin^4\theta=\left(\frac{1-\cos2\theta}2\right)^2=\frac14-\cos2\theta+\frac14\cos^22\theta=\frac14-\cos2\theta+\frac{\cos4\theta+1}8.$$

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

Answer 3

We can solve the integral analytically using standard integration techniques.

First, we can expand the integrand using partial fractions:

$$\frac{x^{-2}(x-\epsilon)(\epsilon+2)^{1.5}(2-\epsilon)^{1.5}}{2.5-2} - \frac{2}{2.5-2} = \frac{2}{3x^2}\left[\frac{(x-2)(\epsilon+2)^{1.5}}{(\epsilon-2)^{0.5}} - \frac{x(\epsilon+2)^{1.5}}{(\epsilon-2)^{0.5}}\right] - \frac{2}{3x^2}$$

Then, we can integrate each term separately:

$$\begin{aligned} &\int \frac{x^{-2}(x-\epsilon)(\epsilon+2)^{1.5}(2-\epsilon)^{1.5}}{2.5-2} - \frac{2}{2.5-2} d\epsilon - x \\ &= -\frac{2}{3x^2} \left[\frac{2(x-2)(\epsilon+2)^{0.5}(\epsilon-2)^{0.5}}{5} - \frac{2x(\epsilon+2)^{0.5}(\epsilon-2)^{0.5}}{3}\right]_{\epsilon=-\infty}^x - x + a \\ &= -\frac{4}{15x}[(x-2)x^{0.5}(2+x)^{0.5} - 4x^{1.5}(2+x)^{0.5}] - x + a \\ &= -\frac{4x^{0.5}}{15}[(x-2)(2+x)^{0.5} - 4x(2+x)^{0.5}] - x + a \end{aligned}$$

Therefore, the solution to the integral equation is:

$$\int_x f(\epsilon) d\epsilon - x = -\frac{4x^{0.5}}{15}[(x-2)(2+x)^{0.5} - 4x(2+x)^{0.5}] - x + a$$

where $f(\epsilon) = \frac{x^{-2}(x-\epsilon)(\epsilon+2)^{1.5}(2-\epsilon)^{1.5}}{2.5-2}$.