Trig substitution integral help needed: int (-3x^5)/sqrt(x^2+2) dx

[unimagineit]

So I have to integrate the following: ∫(-3x⁵)/sqrt(x²+2) dx
To simplify things I always take away the constant and simply add it at the end, so that leaves me with : ∫x⁵/sqrt(x²+2) dx
x=sqrt(2)*tanØ & dx=sqrt(2)sec²Ø
after doing some basic algebra and integration, I end up with 2^5/2(sec⁵Ø/5 - 2sec³Ø/3 + secØ) + C
From here, all I do is replace secØ by sqrt(x²+2)/sqrt(2); because tanØ=x/sqrt(2) ; however I don't end up with the right answer when I do this. According to WolframAlpha I'm doing everything right until I replace secØ by it's value. Can anyone help?

Here's a picture of the integral in case my formatting isn't clear.
https://gyazo.com/51fa251057a7e5d29fd474da78ddd5df

 

[ksdhart]

By all accounts, everything should work out fine. You integrated properly, and the value for \(\displaystyle sec(\theta)\) is the correct value. Thus, the only thing I can conclude is that there must have been an algebra error when you plugged in the value. That said, we cannot provide guidance as to where any errors might be without seeing your work. When you reply back, please include the steps you took when plugging in the value. Thank you.

 

[Prove It]

So I have to integrate the following: ∫(-3x⁵)/sqrt(x²+2) dx
To simplify things I always take away the constant and simply add it at the end, so that leaves me with : ∫x⁵/sqrt(x²+2) dx
x=sqrt(2)*tanØ & dx=sqrt(2)sec²Ø
after doing some basic algebra and integration, I end up with 2^5/2(sec⁵Ø/5 - 2sec³Ø/3 + secØ) + C
From here, all I do is replace secØ by sqrt(x²+2)/sqrt(2); because tanØ=x/sqrt(2) ; however I don't end up with the right answer when I do this. According to WolframAlpha I'm doing everything right until I replace secØ by it's value. Can anyone help?

Here's a picture of the integral in case my formatting isn't clear.
https://gyazo.com/51fa251057a7e5d29fd474da78ddd5df

I think trigonometric substitution will be overkill here...

\(\displaystyle \displaystyle \begin{align*} \int{ -\frac{3\,x^5}{\sqrt{ x^2 + 2 } }\,\mathrm{d}x } &= -\frac{3}{2} \int{ \frac{\left( x^2 + 2 - 2 \right) ^2 }{\sqrt{ x^2 + 2 }} \, 2\,x \, \mathrm{d}x } \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} u = x^2 + 2 \implies \mathrm{d}u = 2\,x\,\mathrm{d}x \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} -\frac{3}{2} \int{ \frac{\left( x^2 + 2 - 2 \right) ^2}{\sqrt{ x^2 + 2 }} \, 2\,x\,\mathrm{d}x } &= -\frac{3}{2} \int{ \frac{\left( u - 2 \right) ^2}{\sqrt{u}}\,\mathrm{d}u } \\ &= -\frac{3}{2} \int{ \frac{u^2 - 4\,u + 4}{u^{\frac{1}{2}}} \,\mathrm{d}u } \\ &= -\frac{3}{2} \int{ \left( u^{\frac{3}{2}} - 4\,u^{\frac{1}{2}} + 4\,u^{-\frac{1}{2}} \right) \,\mathrm{d}u } \end{align*}\)

I'm sure you can go from here...

 

[Deleted member 4993]

I think trigonometric substitution will be overkill here...

\(\displaystyle \displaystyle \begin{align*} \int{ -\frac{3\,x^5}{\sqrt{ x^2 + 2 } }\,\mathrm{d}x } &= -\frac{3}{2} \int{ \frac{\left( x^2 + 2 - 2 \right) ^2 }{\sqrt{ x^2 + 2 }} \, 2\,x \, \mathrm{d}x } \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} u = x^2 + 2 \implies \mathrm{d}u = 2\,x\,\mathrm{d}x \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} -\frac{3}{2} \int{ \frac{\left( x^2 + 2 - 2 \right) ^2}{\sqrt{ x^2 + 2 }} \, 2\,x\,\mathrm{d}x } &= -\frac{3}{2} \int{ \frac{\left( u - 2 \right) ^2}{\sqrt{u}}\,\mathrm{d}u } \\ &= -\frac{3}{2} \int{ \frac{u^2 - 4\,u + 4}{u^{\frac{1}{2}}} \,\mathrm{d}u } \\ &= -\frac{3}{2} \int{ \left( u^{\frac{3}{2}} - 4\,u^{\frac{1}{2}} + 4\,u^{-\frac{1}{2}} \right) \,\mathrm{d}u } \end{align*}\)

I'm sure you can go from here...

That was really neat!!

 

[Berationalgetreal]

Another substitution

I think trigonometric substitution will be overkill here...

\(\displaystyle \displaystyle \begin{align*} \int{ -\frac{3\,x^5}{\sqrt{ x^2 + 2 } }\,\mathrm{d}x } &= -\frac{3}{2} \int{ \frac{\left( x^2 + 2 - 2 \right) ^2 }{\sqrt{ x^2 + 2 }} \, 2\,x \, \mathrm{d}x } \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} u = x^2 + 2 \implies \mathrm{d}u = 2\,x\,\mathrm{d}x \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} -\frac{3}{2} \int{ \frac{\left( x^2 + 2 - 2 \right) ^2}{\sqrt{ x^2 + 2 }} \, 2\,x\,\mathrm{d}x } &= -\frac{3}{2} \int{ \frac{\left( u - 2 \right) ^2}{\sqrt{u}}\,\mathrm{d}u } \\ &= -\frac{3}{2} \int{ \frac{u^2 - 4\,u + 4}{u^{\frac{1}{2}}} \,\mathrm{d}u } \\ &= -\frac{3}{2} \int{ \left( u^{\frac{3}{2}} - 4\,u^{\frac{1}{2}} + 4\,u^{-\frac{1}{2}} \right) \,\mathrm{d}u } \end{align*}\)

I'm sure you can go from here...

Actually there's a better substitution:

\(\displaystyle t\, =\, \sqrt{\strut x^2\, +\, 2\,}\)

Whence

\(\displaystyle dt\, =\, \dfrac{x \ dx}{\sqrt{\strut x^2\, +\, 2\,}}\)

And

\(\displaystyle x^4\, =\, (t^2\, -\, 2)^2\)


With this, we obtain a very simpler integral to calculate:

\(\displaystyle -3\, \int\, \left(\, t^4\, +\, 4\, -\, 4t^2\, \right)\, dt\)