Question

Evaluate integral fraction numerator x to the power of 4 over denominator square root of 4 minus x squared end root end fraction d x

Answer 1

At first, we have the integral

$I=\int \frac{x^4}{\sqrt{4-x^2}}dx$

Then if we use trigonometric substitution

$sin\, \theta =\frac{\sqrt{4-x^2}}{2} \implies \sqrt{4-x^2} = 2\,sin\, \theta \\ cos\, \theta =\frac{x}{2} \\ tan\, \theta =\frac{\sqrt{4-x^2}}{x} \\ x=2\,cos\, \theta \implies dx=-2\,sin\,\theta\,d\theta$

we find that from there we transform the integral to:

$I=\int \frac{x^4}{\sqrt{4-x^2}}dx=\int \frac{(2\,cos\, \theta)^4}{\cancel{2sin\,\theta}}(-\cancel{2\,sin\,\theta}d\theta)$

$\implies I=-2^4\int cos^4\theta \, d\theta$

From there, to solve the trigonometric integral we use the double angle formula for the cosine:

$cos^2\theta=\frac{1}{2}(1 + cos(2\theta)) \\ cos^4\theta=(\frac{1}{2}(1 + cos(2\theta)))^2 \\ cos^4(\theta) = \frac{1}{4}(1 + 2cos(2\theta) + cos^2(2\theta)) \\ cos^4(\theta) = \frac{1}{4}(1 + 2cos(2\theta) + \frac{1}{2}(1 + cos(4\theta)))$

$\implies cos^4(\theta) = \frac{3}{8} + \frac{1}{2}cos(2\theta) + \frac{1}{8}cos(4\theta)$

We solve the integral and we have to use the formulas for the double angle of the sines to then make the proper algebra and find the answer:

$I=-2^4\int cos^4\theta \, d\theta=-2^4\int ( \frac{3}{8} + \frac{1}{2}cos(2\theta) + \frac{1}{8}cos(4\theta))d\theta$

$I=-2^4(\frac{3}{8}\int d\theta + \frac{1}{2}\int cos(2\theta) d\theta + \frac{1}{8}\int cos(4\theta)d\theta)$

$I=-2^4 (\frac{3}{8}\theta + \frac{1}{4}sin(2\theta ) + \frac{1}{32}sin(4\theta ) + C) \\ I=-2^4 (\frac{3}{8}\theta + \frac{1}{4}(2sin\theta cos\theta ) + \frac{1}{32}(2sin(2\theta) cos(2\theta)) + C) \\ I=-2^4 (\frac{3}{8}\theta + \frac{1}{2}sin\theta cos\theta + \frac{1}{16}(2sin\theta cos\theta )(1-2sin^2\theta) + C) \\ I=-2^4 (\frac{3}{8}\theta + \frac{1}{2}sin\theta cos\theta + \frac{1}{16}(2sin\theta cos\theta - 4sin^3\theta cos\theta) + C) \\ I=-2^4 (\frac{3}{8}\theta + \frac{1}{2}sin\theta cos\theta + \frac{1}{8}sin\theta cos\theta - \frac{1}{4}sin^3\theta cos\theta + C)$

$\implies I=-6\theta -10sin\theta cos\theta +4sin^3\theta cos\theta + C$

We only have to substitute to get back to the form with x as the variable to find the integral:

$I=-6\,arc cos (\frac{x}{2}) -\frac{5x\sqrt{4-x^2}}{2} +\frac{x(\sqrt{4-x^2})^3}{4} + C \\ I=-6\,arc cos (\frac{x}{2}) +\sqrt{4-x^2}(-\frac{5x}{2} +\frac{x(4-x^2)}{4} ) + C \\I = -6\,arc cos (\frac{x}{2}) -\sqrt{4-x^2}(\frac{3x}{2} +\frac{x^3}{4} ) + C$

In conclusion, $\int \frac{x^4}{\sqrt{4-x^2}}dx= -6\,arc cos (\frac{x}{2}) -\sqrt{4-x^2}(\frac{3x}{2} +\frac{x^3}{4} ) + C$ .

Reference:

• Varberg, D. E., Purcell, E. J., & Rigdon, S. E. (2007). Calculus with differential equations. Pearson/Prentice Hall.