MHB Completing A Square and Trig Sub

[Nick Cutaia]

I have a problem with the integration of $\int \sqrt{x^2 +4x +5} \,dx$

I first started by completing the square ${x}^{2} +4x + 5 = {x}^{2} +4x +4 - 4 +5 $

After I completed the square the integral became $\int\sqrt{{x}^{2} +4x +4 - 4 +5}\, dx = \int\sqrt{{x+2}^{2}+1} \,dx$

Then I did a trig sub: $\tan\theta = x + 2 \qquad x = \tan\theta - 2 \qquad dx = \sec^2\theta \,d\theta$

Substituting we get: $\int\sqrt{\tan^2\theta+1} \,dx = \sec^3\theta \,d\theta$

And this is where I have been getting stuck. I have forgotten how to integrate $\sec^3\theta$

 

[I like Serena]

Hi Nick Cutaia! Welcome to MHB! (Wave)

Well, I don't remember a $\sec^3$ either...
Instead I suggest a $\sinh$ substitution.

 

[Nick Cutaia]

I assumed an a*tan\theta would be the substitution for a \sqrt{x² + a²} integral.

I'm not following how I would use a sinh substitution.

 

[Dethrone]

Hi Nick Cutaia,

To integrate $\sec^3x$, the trick is to use integration by parts, letting $u=\sec x$ and $dv=\sec^2x$, however, ILS's suggestion of a $\sinh x$ subustitution is most likely better.

 

[Greg]

The sinh substitution definitely leads to a "cleaner" re-substitution of variables.

After the sub you have

$$\int\cosh^2\theta\,\mathrm d\theta$$

Now use integration by parts with $$dv=\cosh\theta\,\mathrm d\theta$$ and $$u=\cosh\theta$$, and the identity $$\sinh^2\theta=\cosh^2\theta-1$$.

Note that $$\dfrac{\mathrm d\cosh\theta}{\mathrm d\theta}=\sinh\theta$$ and $$\dfrac{\mathrm d\sinh\theta}{\mathrm d\theta}=\cosh\theta$$.

 

[Prove It]

The sinh substitution definitely leads to a "cleaner" re-substitution of variables.

After the sub you have

$$\int\cosh^2\theta\,\mathrm d\theta$$

Now use integration by parts with $$dv=\cosh\theta\,\mathrm d\theta$$ and $$u=\cosh\theta$$, and the identity $$\sinh^2\theta=\cosh^2\theta-1$$.

Note that $$\dfrac{\mathrm d\cosh\theta}{\mathrm d\theta}=\sinh\theta$$ and $$\dfrac{\mathrm d\sinh\theta}{\mathrm d\theta}=\cosh\theta$$.

Or even easier, make use of the double angle identities: $\displaystyle \begin{align*} \cosh{(2x)} \equiv 2\cosh^2{(x)} - 1 \end{align*}$ and $\displaystyle \begin{align*} \sinh{(2x)} \equiv 2\sinh{(x)}\cosh{(x)} \end{align*}$ :)

 

[I like Serena]

I assumed an a*tan\theta would be the substitution for a \sqrt{x² + a²} integral.

I'm not following how I would use a sinh substitution.

The way to substitute it is the same: $x+2=\sinh\theta$.
It works because we have $\cosh^2\theta-\sinh^2\theta=1$.

Similarly we should use the $\sin$ substitution when we have $\sqrt{1-x^2}$.