Question

Evaluate the following integrals: $$ \int \frac{x^{2} d x}{\left(4-x^{2}\right)^{5 / 2}} $$

Short answer

The substitution used to solve the integral was x = 2 sin u, which simplifies the denominator and transforms the integral into a trigonometric integral that can be integrated easily.

Step-by-step solution

Choose a suitable substitution

Since the denominator involves the term \((4-x^2)\), we can use the substitution: $$x = 2\sin u$$ This substitution simplifies the denominator. Also, notice that when \(u = 0\), \(x = 0\), which maintains the bounds of the integral if any.

Compute dx and substitute

Now we need to compute \(dx\) in terms of \(du\). Differentiating the substitution equation with respect to \(u\), we have: $$dx = 2\cos u du$$ Now we can substitute \(x = 2\sin u\) and \(dx = 2\cos u du\) back into the integral to transform it into an integral with respect to \(u\).

Transform the integral

Substituting our expressions for \(x\) and \(dx\), we get: $$ \int \frac{(2\sin u)^2 (2\cos u) du}{\left(4-(2\sin u)^2\right)^{5/2}} $$

Simplify and solve the integral

Now, we can simplify the integral further: $$ \int \frac{4\sin^2(u) 2\cos(u) du}{\left(4-4\sin^2(u)\right)^{5/2}} $$ Use the trigonometric identity \(\cos^2(u) = 1- \sin^2(u)\) to rewrite the denominator of the integrand: $$ \int \frac{4\sin^2(u) 2\cos(u) du}{(4\cos^2(u))^{5/2}} $$ Simplify the expression: $$ \int \frac{8\sin^2(u)\cos(u) du}{8\cos^5(u)} $$ Reduce the expression to the following integral: $$ \int \frac{\sin^2(u)}{\cos^4(u)} \cos(u) du $$ Use the trigonometric identity \(\sin^2(u) = 1- \cos^2(u)\) to rewrite the integrand: $$ \int \frac{1 - \cos^2(u)}{\cos^4(u)} \cos(u) du $$ Separate the integral into two parts: $$ \int \left(\frac{1}{\cos^3(u)} - \frac{\cos(u)}{\cos^5(u)}\right) du $$ Now integrate each part separately: $$ \int\sec^3(u) du - \int\sec^5(u) d(u) $$ Compute the integrals by using the standard methods for integrating secant functions: $$ \frac{1}{2}(\sec(u)\tan(u) + \ln|\sec(u) + \tan(u)|) - (\frac{3}{8}\tan(u)\sec(u) + \frac{1}{8}\ln|\sec(u) + \tan(u)|) + C $$ We have now integrated in terms of \(u\). We should return the result to the original variable: \(x\).

Substitute back for x

Use the substitution \(x = 2\sin u\) to find the inverse function: $$ u = \arcsin\left(\frac{x}{2}\right) $$ Substitute the expressions for \(u\) and calculate the corresponding trigonometric functions: $$ \tan(u) = \frac{x}{\sqrt{4-x^2}}, \quad \sec(u) =\frac{2}{\sqrt{4-x^2}} $$ Now, plug these values back into the integral result: $$ \frac{1}{2}\left(\frac{2x}{4-x^2} + \ln\left|\frac{2}{\sqrt{4-x^2}} + \frac{x}{\sqrt{4-x^2}}\right|\right) - \left(\frac{3}{8}\frac{x}{\sqrt{4-x^2}}\frac{2}{\sqrt{4-x^2}} + \frac{1}{8}\ln\left|\frac{2}{\sqrt{4-x^2}} + \frac{x}{\sqrt{4-x^2}}\right|\right) + C $$ Combining the log terms and simplifying: $$ \frac{1}{8}\left(8\frac{x}{4-x^2}+\ln\left|\frac{2+x}{\sqrt{4-x^2}}\right|\right)-\frac{3}{4}\cdot\frac{x^2}{4-x^2} + C $$ And that's our final result: $$ \int \frac{x^2 dx}{(4-x^2)^{5/2}}=\frac{1}{8}\left(8\frac{x}{4-x^2}+\ln\left|\frac{2+x}{\sqrt{4-x^2}}\right|\right)-\frac{3}{4}\cdot\frac{x^2}{4-x^2} + C $$