How do you integrate #int(dx/(sqrt(2x^3 + 2x + 5)))#?

Question

Cameron Alexander · Accepted Answer

The integral you wrote is impossible to do using "elementary functions" (at least according to the computer program Mathematica...it requires "Elliptic Functions").

I think you probably meant to do #\int(\frac{dx}{\sqrt{2x^{2}+2x+5}})#. The first steps in doing this integral is to use a bit of tricky algebra (including the technique of completing the square) to get:

#\int\frac{dx}{\sqrt{2x^{2}+2x+5}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{x^{2}+x+\frac{5}{2}}}#

#=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{x^{2}+x+\frac{1}{4}+\frac{9}{4}}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{(x+\frac{1}{2})^{2}+\frac{9}{4}}}#.

Now do the trigonometric substitution #x+\frac{1}{2}=\frac{3}{2}\tan(\theta)# do so that #\theta=\arctan(\frac{2}{3}x+\frac{1}{3})# and #dx=\frac{3}{2}\sec^{2}(\theta)d\theta#.

Then the integral becomes:

#\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{(x+\frac{1}{2})^{2}+\frac{9}{4}}}=\frac{1}{\sqrt{2}}\int\frac{\frac{3}{2}\sec^{2}(\theta)d\theta}{\sqrt{\frac{9}{4}(\tan^{2}(\theta)+1)}}#, which simplifies to

#\frac{1}{\sqrt{2}}\int\frac{\sec^{2}(\theta)d\theta}{\sqrt{\sec^{2}(\theta)}}#

after using a trigonometric identity. Technically-speaking, #\sqrt{\sec^{2}(\theta)}=|sec(\theta)|#, however, it is fine to get rid of the absolute value signs for the purpose of finishing the integral (assume #\sec(\theta)\geq 0#).

Hence, the integral becomes #\frac{1}{\sqrt{2}}\int\sec(\theta)d\theta#. This last integral looks simple, but it is still tricky to do. One trick that works is to multiply it by #\frac{\sec(\theta)+\tan(\theta}}{\sec(\theta)+\tan(\theta)# and then do a substitution (you can try that if you want). Wolfram Alpha gives the answer as #\frac{1}{\sqrt{2}}(ln(cos(\theta/2)+sin(\theta/2))-\ln(\cos(\theta/2)-\sin(\theta/2)))+C#. Now just replace #\theta# by #\theta=\arctan(\frac{2}{3}x+\frac{1}{3})# and you are done:

#\int(\frac{dx}{\sqrt{2x^{2}+2x+5}})=\frac{1}{\sqrt{2}}(ln(cos(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))+sin(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3})))-\ln(\cos(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))-\sin(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))))+C#

It is possible to simplify this by using half angle formulas and the following equations, but it may not be worth the trouble: #\cos(\arctan(\frac{2}{3}x+\frac{1}{3}))=\frac{1}{\sqrt{1+(\frac{2}{3}x+\frac{1}{3})^{2}}}=\frac{3}{\sqrt{4x^{2}+4x+10}}#

#\sin(\arctan(\frac{2}{3}x+\frac{1}{3}))=\frac{\frac{2}{3}x+\frac{1}{3}}{\sqrt{1+(\frac{2}{3}x+\frac{1}{3})^{2}}}=\frac{2x+1}{\sqrt{4x^{2}+4x+10}}#

Cameron Alexander · Answer

undefined To integrate ∫dx/(√(2x^3 + 2x + 5)), we can use the method of trigonometric substitution. Let's make the substitution x = (5/2)sin^2(θ). Then dx = 5sin(2θ)dθ.

Substituting into the integral, we get:

∫dx/(√(2x^3 + 2x + 5)) = ∫5sin(2θ)dθ/(√(2(5/2)sin^2(θ)^3 + 2(5/2)sin^2(θ) + 5))
                          = ∫5sin(2θ)dθ/(√(5sin^2(θ)^3 + 5sin^2(θ) + 5))

Now, let's simplify the expression inside the square root:

5sin^2(θ)^3 + 5sin^2(θ) + 5 = 5(sin^2(θ)^2 + sin^2(θ) + 1)
                              = 5(sin^4(θ) + sin^2(θ) + 1)

Now, we can rewrite the integral as:

∫5sin(2θ)dθ/(√(5(sin^4(θ) + sin^2(θ) + 1)))

Factoring out a sin^2(θ) from the square root:

= ∫5sin(2θ)dθ/(√(5sin^2(θ)(sin^2(θ)^2 + 1 + sin^2(θ))))

= ∫5sin(2θ)dθ/(√(5sin^2(θ)(sin^2(θ) + 1)))

Notice that sin(2θ) = 2sin(θ)cos(θ), so we can rewrite the integral as:

= ∫(10sin(θ)cos(θ)dθ/(√(5sin^2(θ)(sin^2(θ) + 1))))

Using the identity sin^2(θ) + cos^2(θ) = 1, we have cos^2(θ) = 1 - sin^2(θ). Substituting this into the integral:

= ∫(10sin(θ)cos(θ)dθ/(√(5sin^2(θ)(1 - sin^2(θ) + 1))))

= ∫(10sin(θ)cos(θ)dθ/(√(5sin^2(θ)(2 - sin^2(θ)))))

Now, let u = sin(θ), then du = cos(θ)dθ:

= ∫(10u(1 - u^2)dθ/(√(5u^2(2 - u^2))))

= ∫(10u(1 - u^2)du/(√(10u^2 - 5u^4)))

Now we can use a standard trigonometric integral to solve this. After integrating, remember to substitute back for θ in terms of x.