Integrating Inverse Trigonometric Functions with Substitution

[iRaid]

Homework Statement

$$\int \frac{dx}{(x^{2}+2x+2)^{2}}$$

Homework Equations

The Attempt at a Solution

I'm just going to skip down to where something is wrong (the beginning is a u-substitution and trig. substitution so I don't want to type all that out)
After all the trig/u-subs I end up with:
$$\int cos^{2}\theta d \theta=\int \frac{1}{2}+\frac{cos2 \theta}{2}d \theta=\frac{\theta}{2}+\frac{sin2\theta}{4}+C$$

So now I substitute back, u=tanθ:
$$\frac{tan^{-1}u}{2}+\frac{u}{2\sqrt{u^{2}+1}}$$
and u=x+1:
I THINK THE ERROR IS HERE SOMEWHERE, BUT NOT SURE WHERE
$$\frac{1}{2}(tan^{-1}(x+1)+\frac{x+1}{\sqrt{(x+1)^{2}+1}})+C$$Thank you

 

[Karnage1993]

Your error is when you substitute $\theta$ back into $\displaystyle \frac{sin2\theta}{4} \not = \frac{u}{2\sqrt{u^{2}+1}}$

 

[iRaid]

If you draw the triangle where u=tanθ, then sine of that would be $\frac{u}{\sqrt{u^{2}+1}}$ wouldn't it? And then multiplying that by 2 would just be 2 times that, but I think that's the error, I'm not sure what the 2 does to the sine function.

 

[Karnage1993]

If you draw the triangle where u=tanθ, then sine of that would be $\frac{u}{\sqrt{u^{2}+1}}$ wouldn't it? And then multiplying that by 2 would just be 2 times that, but I think that's the error, I'm not sure what the 2 does to the sine function.

No, $sin(2arctan(x)) \not = 2sin(arctan(x))$. I don't know how to derive $sin(2arctan(x))$ but a Google search could help.

Instead of finding an expression without trig functions, sometimes it's just best to leave it as is. The answer is still perfectly valid.

 

[iRaid]

Ok yeah I see that's the problem, but even after googling it, that doesn't help :\

 

[iRaid]

Still don't understand this, could someone explain how to get sin(2arctan(x))?

 

[Mute]

Use $\sin(2\theta) = 2\sin \theta \cos \theta$, then write the sin and cos in terms of u.

 

[iRaid]

Use $\sin(2\theta) = 2\sin \theta \cos \theta$, then write the sin and cos in terms of u.

Wow forgot that, thanks so much.