16. find the indefinite integral by making the substitution. $x = 5sin\theta$ \n$int\frac{x}{(25…

Explanation:

Step1: Find $\mathrm{d}x$

Given $x = 5\sin\theta$, then $\mathrm{d}x=5\cos\theta\mathrm{d}\theta$. Also, $25 - x^{2}=25-25\sin^{2}\theta = 25(1 - \sin^{2}\theta)=25\cos^{2}\theta$.

Step2: Substitute $x$ and $\mathrm{d}x$ into the integral

The integral $\int\frac{x}{(25 - x^{2})^{\frac{3}{2}}}\mathrm{d}x$ becomes $\int\frac{5\sin\theta}{(25\cos^{2}\theta)^{\frac{3}{2}}}\cdot5\cos\theta\mathrm{d}\theta=\int\frac{5\sin\theta}{125\cos^{3}\theta}\cdot5\cos\theta\mathrm{d}\theta=\int\frac{\sin\theta}{\cos^{2}\theta}\mathrm{d}\theta$.

Step3: Use substitution for new - integral

Let $u = \cos\theta$, then $\mathrm{d}u=-\sin\theta\mathrm{d}\theta$. The integral $\int\frac{\sin\theta}{\cos^{2}\theta}\mathrm{d}\theta=-\int\frac{\mathrm{d}u}{u^{2}}$.

Step4: Integrate $-\int\frac{\mathrm{d}u}{u^{2}}$

We know that $\int u^{n}\mathrm{d}u=\frac{u^{n + 1}}{n+1}+C$ for $n\neq - 1$. Here $n=-2$, so $-\int\frac{\mathrm{d}u}{u^{2}}=-\int u^{-2}\mathrm{d}u=\frac{1}{u}+C$.

Step5: Back - substitute $u$ and $\theta$

Since $u = \cos\theta$ and $\cos\theta=\frac{\sqrt{25 - x^{2}}}{5}$, then $\frac{1}{u}=\frac{1}{\cos\theta}=\frac{1}{\frac{\sqrt{25 - x^{2}}}{5}}\cdot\frac{1}{5}=\frac{1}{\sqrt{25 - x^{2}}}+C$.

Answer:

A. $\frac{1}{\sqrt{25 - x^{2}}}+C$