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)^{((3)/(2))}}dx`

using a trigonometric substitution

Let

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

substitute in the original expression

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

Remember that

`\tan ^2u+1=\sec ^2u`
`\int \frac{\sec^2u\cdot\tan^2u}{27(\tan^2u+1)^{((3)/(2))}}du=(1)/(27)\int \frac{\tan ^2u}{\sec u^{}}du`
`(1)/(27)\int \frac{\tan^2u}{\sec u^{}}du=(1)/(27)\int (\cos u\cdot\tan ^2u)^{}du`
`(1)/(27)\int (\cos u\cdot\tan ^2u)^{}du=(1)/(27)\int \cos u\cdot(\sec ^2-1)^{}du`
`(1)/(27)\int \cos u\cdot(\sec ^2-1)^{}du=(1)/(27)\int (\sec u-\cos u)^{}du`
`(1)/(27)\int (\sec u-\cos u)^{}du=(1)/(27)\int \sec u^{}du-(1)/(27)\int \cos u^{}du`
`(1)/(27)\int \sec u^{}du-(1)/(27)\int \cos u^{}du=(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

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

simplify

`(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`