Question

$\int\frac{dx}{(49 - x^{2})^{3/2}}$

Explanation:

Step1: Use trigonometric - substitution

Let $x = 7\sin\theta$, then $dx=7\cos\theta d\theta$. And $49 - x^{2}=49 - 49\sin^{2}\theta=49\cos^{2}\theta$.

Step2: Rewrite the integral

The integral $\int\frac{dx}{(49 - x^{2})^{\frac{3}{2}}}$ becomes $\int\frac{7\cos\theta d\theta}{(49\cos^{2}\theta)^{\frac{3}{2}}}=\int\frac{7\cos\theta d\theta}{49^{\frac{3}{2}}\cos^{3}\theta}=\int\frac{7\cos\theta d\theta}{343\cos^{3}\theta}=\frac{1}{49}\int\frac{d\theta}{\cos^{2}\theta}$.

Step3: Integrate $\frac{1}{\cos^{2}\theta}$

Since $\int\frac{d\theta}{\cos^{2}\theta}=\tan\theta + C$. So $\frac{1}{49}\int\frac{d\theta}{\cos^{2}\theta}=\frac{1}{49}\tan\theta + C$.

Step4: Substitute back $\theta$ in terms of $x$

Since $x = 7\sin\theta$, then $\sin\theta=\frac{x}{7}$ and $\cos\theta=\frac{\sqrt{49 - x^{2}}}{7}$, and $\tan\theta=\frac{x}{\sqrt{49 - x^{2}}}$.

Answer:

$\frac{x}{49\sqrt{49 - x^{2}}}+C$