integration by trigonometric substitution: x^2/sqrt(x^2+6)

[T_TEngineer_AdamT_T]

\(\displaystyle \L \int \frac{x^2}{\sqrt{x^2+6}}dx\)

Let \(\displaystyle \L x = \sqrt{6}\tan{\theta}\)

\(\displaystyle \L dx = \sqrt{6}\sec^{2}{\theta}\)

\(\displaystyle \L \int \frac{6\tan^2{\theta}(\sqrt{6}\sec^2{\theta})}{\sqrt{6\sec^2{\theta})\)

\(\displaystyle \L \int 6\tan^2{\theta}\sec{\theta}d\theta\)

\(\displaystyle \L 6\int (\sec^2{\theta} - 1)\sec{\theta}d{\theta}\)

\(\displaystyle \L 6\int (\sec^3{\theta} - \sec{\theta})d{\theta}\)

\(\displaystyle \L 6\int (\sec^3{\theta}d{\theta} - 6ln|\sec{\theta} + \tan{\theta}|\)

Integrating by parts for \(\displaystyle \L sec^3{theta}\), this becomes:

\(\displaystyle \L \int \sec^3{\theta} = 3\sec{\theta}\tan{\theta} + \frac{1}{2}ln|\sec{\theta} + \tan{\theta}|\)

To continue:

\(\displaystyle \L =3\sec{\theta}\tan{\theta} + \frac{1}{2}ln|\sec{\theta} + \tan{\theta}|-6ln|\sec{\theta} + \tan{\theta}|\)

\(\displaystyle \L =3\sec{\theta}\tan{\theta} - \frac{11}{2}ln|\sec{\theta}+\tan{\theta}|\)

I draw a triangle and obtain the values of x. This becomes:

\(\displaystyle \L \frac{1}{2}x{\sqrt{x^2+6}} - \frac{11}{2}ln|\frac{\sqrt{x^2+6}+x}{\sqrt{6}}|\)

Is my answer correct?

 

[galactus]

You're OK down to \(\displaystyle \L\\6\int[sec^{3}{\theta}-sec{\theta}]d{\theta}\)

I believe there may be a slight error at the 11/2. Should be 6/2. You can see it easier than I.

\(\displaystyle \L\\6\int[sec^{3}{\theta}-sec{\theta}]d{\theta}=6\left(\frac{1}{2}sec{\theta}tan{\theta}-\frac{1}{2}ln|sec{\theta}+tan{\theta}|\right)+C_{1}\)

=\(\displaystyle \L\\\frac{1}{2}x\sqrt{6+x^{2}}-\frac{6}{2}ln(\frac{\sqrt{6+x^{2}}+x}{\sqrt{6}})=\frac{1}{2}x\sqrt{6-x^{2}}-3ln(\sqrt{6+x^{2}}+x)+C\)

 

[T_TEngineer_AdamT_T]

thanks galactus:
but im still confused:

u integrate first sec^3{\theta} right?
\(\displaystyle \L \int \sec^3{\theta} d{\theta} - \int{\sec{\theta}}\)

now sec^3{theta} is integrating by parts? and 6sec {theta} can be integrated to 6ln|sec{theta} + tan{theta}|?

so the value of \int sec^3{theta} is added up to 6ln|sec{theta} + tan{theta}|?

 

[Deleted member 4993]

The value of

[int]sec^3(x) dx = 1/2 *{sec(x)*tan(x) + [int]sec(x) dx}


so

[int]{sec^3(x) - sec(x) } dx = 1/2 *{sec(x)*tan(x) + [int]sec(x) dx} - [int]sec(x) dx


[int]{sec^3(x) - sec(x) } dx = 1/2 *{sec(x)*tan(x) - [int]sec(x) dx}

6*[int]{sec^3(x) - sec(x) } dx = 3 *{sec(x)*tan(x) - 3*[int]sec(x) dx}

 

[T_TEngineer_AdamT_T]

ok thanks

 

[galactus]

\(\displaystyle \L\\\int{sec^{3}(\theta)}dx=\)

=\(\displaystyle \L\\\frac{sec(\theta)tan(\theta)}{2}+\frac{1}{2}ln|sec(\theta)+tan(\theta)|\)

\(\displaystyle \L\\\int{sec(\theta)dx=ln|sec(\theta)+tan(\theta)|+C\)

Now, write it in terms of x if you wish.

\(\displaystyle \L\\sec{\theta}=\frac{\sqrt{x^{2}+6}}{\sqrt{6}}\)

\(\displaystyle \L\\tan{\theta}=\frac{x}{\sqrt{6}}\)


Here's the general formula:

\(\displaystyle \L\\\int\frac{x^{2}}{\sqrt{x^{2}+a^{2}}}=\frac{x}{2}\sqrt{x^{2}+a^{2}}-\frac{a^{2}}{2}ln|x+\sqrt{x^{2}+a^{2}}|+C\)