Seemingly simple integral gets complicated

[EbolaPox]

Well, I'm an AP Calculus BC student, and I always liked to get ahead, so before the year started, I went through Calc I and II. Now that I'm in the class with a 100+, I'm simply refreshing myself on the Calc II portion. My brother decided to give me a seemingly simple problem that turned out more complicated than I expected. Any assistance would be greatly appreciated.
Problem: Evaluate integral
$$\int \sin{x} \sqrt{(\cos{x})^2 + 1}dx$$
I decided to use substitution initially, setting u = cos x. dx = du/-sin(x), thus eliminating the sin(x) and leaving $$\int \sqrt{u^2 + 1}du$$
I then thought my best bet would be to approach this with trignometric substitution. I said tan($$\theta$$) = u so
$$\sec{\theta}^2d\theta = du$$
$$\int (\sec{\theta})^2\sqrt{(\tan(\theta)} +1)dx$$
I then evaluated $$\sqrt{ (\tan{\theta})^2 +1 } = (\sec{\theta})$$
Thus, my integral was simplified (relatively) down to
$$\int (\sec{\theta})^3 d\theta$$
It just gets worse from there. Am I making some horrid mistake in my trignometric substitution or before or have I completely approached this incorrectly? I have more work that I've done, but I want to ensure that this is correct so far. Any help would be great. Thanks

 

[Fermat]

try u = sinht

 

[*melinda*]

It looks a-ok so far. Just don't forget the minus sign in your first u-substitution.

Do you know how to integrate secant cubed?

 

[HallsofIvy]

Your brother is a smart-ass!

To integrate $sec^3(\theta)d(\theta)$, first write it as $\frac{d\theta}{cos^3(\theta)}$. Now multiply numerator and denominator by $cos(\theta)$ to get $\frac{cos(\theta)d\theta}{cos^4(\theta)}= \frac{cos(\theta)d\theta}{(1-sin^2(\theta)^2}$.
Let $y= sin(\theta)$ and this becomes $\frac{dy}{(1-y^2)^2}$ which can be done by partial fractions.

 

[*melinda*]

When I was in calculus I was taught that the only way to integrate secant cubed was by parts. Obviously this is not the case .

By parts does not take too long, if you remember what to choose for u and dV!

 

[amcavoy]

To elaborate on what Fermat said, the integral can be written as:

$$\int\sqrt{1+x^2}\,dx$$

Then you can use the suggested substitution.

And speaking of parts, try using dv = sec²θ.

 

[EbolaPox]

Thanks to everyone who replied. For some reason, I didn't think about attempting integration by parts.

Thanks to suggestsions everyone, I'll try them out. A question though, for the y = sinh(t), if I'm correct, is that hyperbolic trignometric substitution? I'm not very familiar with hyperbolic functions. I'll read up on it. Thanks to everyone. Once I get some more work done on it, I'll post again to see if I did it correctly.

 

[Fermat]

... A question though, for the y = sinh(t), if I'm correct, is that hyperbolic trignometric substitution? ...

Yes, that's a hyperbolic substitution. I goes like this,

$$\int\ \sqrt{u^2 + 1}\ du$$

$$\mbox{let } u = sinh\ t\mbox{, then,}$$

$$du = cosh\ t\ dt$$

carrying out the substitution,

$$\int\ \sqrt{sinh^2\ t + 1}\cdot cosh\ t\ dt$$
$$\int\ \sqrt{cosh^2\ t}\cdot cosh\ t\ dt$$
$$\int\ cosh^2\ t\ dt$$

$$\mbox{There is an ordinary trig identity, } cos^2 x = \frac{1}{2} (1 + cos2x)$$
all you have to do now is find the equivalent hyperbolic trig identity (try google), substitute, integrate, and back-substitute for x in the original integral.

 

[EbolaPox]

Fascinating: I've never seen hyperbolic trignometry used for substitution. Thank you for elucidating on that. I''ll go study up on that.