Question

Evaluate the indefinite integral, using a trigonometric substitution and a triangle to express the answer in terms of x. Use trig substitution to fully convert the integral to a 0-integral. You do not have to compute the 0-integral.

Answer 1

we have the expression

$\int \frac{x^2}{(1+9x^2)^{(\frac{3}{2})}}dx$

using a trigonometric substitution

Let

$\begin{gathered} x=\frac{\tan u}{3} \\ u=\arctan (3x) \\ dx=\frac{\sec ^2u}{3}du \end{gathered}$

substitute in the original expression

$\int \frac{\sec ^2u\cdot\tan ^2u}{27(\tan ^2u+1)^{(\frac{3}{2})}}du$

Remember that

$\tan ^2u+1=\sec ^2u$$\int \frac{\sec^2u\cdot\tan^2u}{27(\tan^2u+1)^{(\frac{3}{2})}}du=\frac{1}{27}\int \frac{\tan ^2u}{\sec u^{}}du$$\frac{1}{27}\int \frac{\tan^2u}{\sec u^{}}du=\frac{1}{27}\int (\cos u\cdot\tan ^2u)^{}du$$\frac{1}{27}\int (\cos u\cdot\tan ^2u)^{}du=\frac{1}{27}\int \cos u\cdot(\sec ^2-1)^{}du$$\frac{1}{27}\int \cos u\cdot(\sec ^2-1)^{}du=\frac{1}{27}\int (\sec u-\cos u)^{}du$$\frac{1}{27}\int (\sec u-\cos u)^{}du=\frac{1}{27}\int \sec u^{}du-\frac{1}{27}\int \cos u^{}du$$\frac{1}{27}\int \sec u^{}du-\frac{1}{27}\int \cos u^{}du=\frac{1}{27}\lbrack\ln (\tan u+\sec u)-\sin u\rbrack$

Remember that

$u=\arctan (3x)$$\tan (\arctan (3x))=3x$

using the triangle

Find out the value of H

Applying the Pythagorean Theorem

H^2=(3x)^2+1^2

H^2=9x^2+1

H=√(9x^2+1)

$\begin{gathered} \sin u=\frac{3x}{\sqrt[]{9x^2+1}} \\ \sec u=\sqrt[]{9x^2+1} \end{gathered}$

substitute

$\frac{1}{27}\lbrack\ln (\tan u+\sec u)-\sin u\rbrack=\frac{1}{27}\lbrack\ln (3x+\sqrt[]{9x^2+1})-\frac{3x}{\sqrt[]{9x^2+1}}\rbrack$

simplify

$\frac{1}{27}\lbrack\ln (3x+\sqrt[]{9x^2+1})-\frac{3x}{\sqrt[]{9x^2+1}}\rbrack=\frac{\ln (3x+\sqrt[]{9x^2+1})}{27}-\frac{x}{9\sqrt[]{9x^2+1}}+C$

therefore

the answer is

$\frac{\ln(3x+\sqrt[]{9x^2+1})}{27}-\frac{x}{9\sqrt[]{9x^2+1}}+C$