Find $\int \frac{2x^3}{1+x^4} \text{ d}x$. Find $\int \cos 4x \sin 10x \text{ d}x$. Show, by means of the substitution $x = \tan \theta$, that $\int_{0}^{1} \frac{x^3}{\left(x^2+1\right)^3} \text{ d}x = \int_{0}^{\frac{\pi}{4}} \cos \theta \sin^3 \theta \text{ d}\theta.$ Hence find the exact value of $\int_{0}^{1} \frac{x^3}{\left(x^2+1\right)^3} \text{ d}x$.

1. Relate the numerator to the derivative of the denominator by extracting a constant factor

$\int \frac{2x^3}{1+x^4} \text{ d}x = \frac{1}{2} \int \frac{4x^3}{1+x^4} \text{ d}x$

2. Integrate to obtain the natural logarithm, including the constant of integration

$= \frac{1}{2} \ln(1+x^4) + c$

1. Express the product of trigonometric functions as a sum using the factor formula.

Using the identity $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$, let $A = 10x$ and $B = 4x$:

$\cos 4x \sin 10x = \frac{1}{2}[\sin(10x+4x) + \sin(10x-4x)]$
$= \frac{1}{2}(\sin 14x + \sin 6x)$

2. Integrate the resulting expression with respect to $x$.

$\int \cos 4x \sin 10x \text{ d}x = \int \frac{1}{2}(\sin 14x + \sin 6x) \text{ d}x$
$= \frac{1}{2} \left( -\frac{\cos 14x}{14} - \frac{\cos 6x}{6} \right) + C$
$= -\frac{1}{28}\cos 14x - \frac{1}{12}\cos 6x + C$

1. Find the differential $\text{d}x$ and the new limits in terms of $\theta$.

Let $x = \tan \theta$.
Differentiating with respect to $\theta$:
$\frac{\text{d}x}{\text{d}\theta} = \sec^2 \theta \implies \text{d}x = \sec^2 \theta \text{ d}\theta$

Change of limits:
When $x = 0$, $\tan \theta = 0 \implies \theta = 0$
When $x = 1$, $\tan \theta = 1 \implies \theta = \frac{\pi}{4}$

2. Substitute $x$, $\text{d}x$, and the limits into the integral, and apply the Pythagorean identity.

Substituting into the integral:
$\int_{0}^{1} \frac{x^3}{\left(x^2+1\right)^3} \text{ d}x = \int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\left(\tan^2 \theta + 1\right)^3} \sec^2 \theta \text{ d}\theta$

Using the identity $\tan^2 \theta + 1 = \sec^2 \theta$:
$= \int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\left(\sec^2 \theta\right)^3} \sec^2 \theta \text{ d}\theta$
$= \int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sec^6 \theta} \sec^2 \theta \text{ d}\theta$
$= \int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sec^4 \theta} \text{ d}\theta$

3. Simplify the trigonometric expression to show the target integrand.

Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$:
$= \int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin^3 \theta}{\cos^3 \theta}}{\frac{1}{\cos^4 \theta}} \text{ d}\theta$
$= \int_{0}^{\frac{\pi}{4}} \frac{\sin^3 \theta}{\cos^3 \theta} \times \cos^4 \theta \text{ d}\theta$
$= \int_{0}^{\frac{\pi}{4}} \cos \theta \sin^3 \theta \text{ d}\theta \quad \text{(shown)}$

4. Integrate the simplified expression using the reverse chain rule.

To evaluate the integral, we can recognise it is of the form $\int [\mathrm{f}(\theta)]^n \mathrm{f}'(\theta) \text{ d}\theta$, since $\frac{\text{d}}{\text{d}\theta}(\sin \theta) = \cos \theta$:
$\int_{0}^{\frac{\pi}{4}} \sin^3 \theta \cos \theta \text{ d}\theta = \left[ \frac{\sin^4 \theta}{4} \right]_0^{\frac{\pi}{4}}$

5. Evaluate the integral at the upper and lower limits to find the exact value.

Substituting the limits:
$= \frac{1}{4} \left( \sin \frac{\pi}{4} \right)^4 - \frac{1}{4} \left( \sin 0 \right)^4$
$= \frac{1}{4} \left( \frac{1}{\sqrt{2}} \right)^4 - 0$
$= \frac{1}{4} \left( \frac{1}{4} \right) = \frac{1}{16}$