Question

evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.)
int\frac{x^{3}}{sqrt{x^{2}+100}}dx,quad x = 10\tan(\theta)
sketch and label the associated right - triangle.

Explanation:

Step1: Find $dx$

Given $x = 10\tan(\theta)$, then $dx=10\sec^{2}(\theta)d\theta$. Also, $\sqrt{x^{2}+100}=\sqrt{100\tan^{2}(\theta)+100}=10\sec(\theta)$.

Step2: Substitute into the integral

The integral $\int\frac{x^{3}}{\sqrt{x^{2}+100}}dx$ becomes $\int\frac{(10\tan(\theta))^{3}}{10\sec(\theta)}\cdot10\sec^{2}(\theta)d\theta=\int1000\tan^{3}(\theta)\sec(\theta)d\theta$.
Since $\tan^{2}(\theta)=\sec^{2}(\theta) - 1$, we have $\int1000\tan^{3}(\theta)\sec(\theta)d\theta=1000\int\tan^{2}(\theta)\tan(\theta)\sec(\theta)d\theta=1000\int(\sec^{2}(\theta)-1)\tan(\theta)\sec(\theta)d\theta$.
Let $u = \sec(\theta)$, then $du=\tan(\theta)\sec(\theta)d\theta$.
The integral is $1000\int(u^{2}-1)du$.

Step3: Integrate with respect to $u$

$1000\int(u^{2}-1)du=1000(\frac{u^{3}}{3}-u)+C$.

Step4: Substitute back $u = \sec(\theta)$

$1000(\frac{\sec^{3}(\theta)}{3}-\sec(\theta))+C$.
From $x = 10\tan(\theta)$, we know $\tan(\theta)=\frac{x}{10}$, and $\sec(\theta)=\frac{\sqrt{x^{2}+100}}{10}$.
So the integral is $1000[\frac{(\frac{\sqrt{x^{2}+100}}{10})^{3}}{3}-\frac{\sqrt{x^{2}+100}}{10}]+C=\frac{1}{3}(x^{2}+100)^{\frac{3}{2}}- 100\sqrt{x^{2}+100}+C$.

For the right - triangle:
If $x = 10\tan(\theta)$, then $\tan(\theta)=\frac{x}{10}$. In a right - triangle, if the opposite side to the angle $\theta$ is $x$ and the adjacent side is $10$, by the Pythagorean theorem, the hypotenuse is $\sqrt{x^{2}+100}$. So the correct right - triangle is the one with adjacent side $10$, opposite side $x$ and hypotenuse $\sqrt{x^{2}+100}$.

Answer:

$\frac{1}{3}(x^{2}+100)^{\frac{3}{2}}-100\sqrt{x^{2}+100}+C$, and the first right - triangle (adjacent side $10$, opposite side $x$, hypotenuse $\sqrt{x^{2}+100}$) is correct.