use the accompanying sections of a table of integrals to evaluate the following indefinite integral. the…

use the accompanying sections of a table of integrals to evaluate the following indefinite integral. the integral may require preliminary work, such as completing the square or changing variables, before it can be found in a table.\n int\frac{dx}{(49 + 25x^{2})^{\frac{3}{2}}}\nclick here to view basic integrals. click here to view trigonometric integrals. click here to view reduction formulas for trigonometric functions. click here to view integrals involving squares of (x) and (a). click here to view integrals involving (axpm b). click here to view other integrals.\n int\frac{dx}{(49 + 25x^{2})^{\frac{3}{2}}}=square

use the accompanying sections of a table of integrals to evaluate the following indefinite integral. the integral may require preliminary work, such as completing the square or changing variables, before it can be found in a table.\n int\frac{dx}{(49 + 25x^{2})^{\frac{3}{2}}}\nclick here to view basic integrals. click here to view trigonometric integrals. click here to view reduction formulas for trigonometric functions. click here to view integrals involving squares of (x) and (a). click here to view integrals involving (axpm b). click here to view other integrals.\n int\frac{dx}{(49 + 25x^{2})^{\frac{3}{2}}}=square

Answer

Explanation:

Step1: Rewrite the denominator

First, rewrite $49 + 25x^{2}$ as $49\left(1+\frac{25}{49}x^{2}\right)=49\left(1 + \left(\frac{5}{7}x\right)^{2}\right)$. So the integral $\int\frac{dx}{(49 + 25x^{2})^{\frac{3}{2}}}=\int\frac{dx}{49^{\frac{3}{2}}\left(1+\left(\frac{5}{7}x\right)^{2}\right)^{\frac{3}{2}}}=\frac{1}{343}\int\frac{dx}{\left(1+\left(\frac{5}{7}x\right)^{2}\right)^{\frac{3}{2}}}$.

Step2: Use substitution

Let $u=\frac{5}{7}x$, then $du=\frac{5}{7}dx$ and $dx=\frac{7}{5}du$. The integral becomes $\frac{1}{343}\cdot\frac{7}{5}\int\frac{du}{(1 + u^{2})^{\frac{3}{2}}}$.

Step3: Use trig - substitution

Let $u = \tan\theta$, then $du=\sec^{2}\theta d\theta$ and $1 + u^{2}=1+\tan^{2}\theta=\sec^{2}\theta$. The integral $\frac{1}{343}\cdot\frac{7}{5}\int\frac{du}{(1 + u^{2})^{\frac{3}{2}}}=\frac{1}{245}\int\frac{\sec^{2}\theta d\theta}{(\sec^{2}\theta)^{\frac{3}{2}}}=\frac{1}{245}\int\frac{\sec^{2}\theta d\theta}{\sec^{3}\theta}=\frac{1}{245}\int\cos\theta d\theta$.

Step4: Integrate

Integrating $\frac{1}{245}\int\cos\theta d\theta=\frac{1}{245}\sin\theta + C$.

Step5: Back - substitute

Since $u = \tan\theta$, we have $\sin\theta=\frac{u}{\sqrt{1 + u^{2}}}$. Substituting back $u=\frac{5}{7}x$, we get $\frac{1}{245}\cdot\frac{\frac{5}{7}x}{\sqrt{1+\left(\frac{5}{7}x\right)^{2}}}+C=\frac{x}{343\sqrt{49 + 25x^{2}}}+C$.

Answer:

$\frac{x}{343\sqrt{49 + 25x^{2}}}+C$