Integration using Trig Substitution

[Brickster]

Homework Statement

$$\int \frac{cosx dx}{\sqrt{1 + sin^{2}x}}$$

Homework Equations

Expression: $$\sqrt{a^{2} + x^{2}}$$
Substitution: x = a*tan$$\Theta$$
Identity: 1 + tan$$^{2}\Theta = sec^{2}\Theta$$

The Attempt at a Solution

I have tried using Trig Substitution, but I end up getting an equation much like the one I started only it contains secants and tangents instead of cosine and sine. For some reason I only see a circular route that would just take me back to the original equation.

 

[cristo]

Try the substitution y=sin(x). That should then put the integrand in a familiar form.

 

[Brickster]

as in changing the bottom from 1 + (sinx)^2 to 1 + y^2 ?I can get to:

$$\int \frac{sec^{2}\Theta d\Theta}{\sqrt{1+tan^{2}\Theta}}$$

But after that it makes no sense, I just get back to the original equation when I sub. again.

 

[Gib Z]

After y=sin x, and then a tan theta = y, you should have the same integrand with a plus sign, not a minus. Then use the Pythagorean identities to simplify the square root.

 

[Brickster]

So I should end up with:

$$\int sec\Theta d\Theta}$$

Correct?

then I would get ln $$\left| sec\Theta+tan\Theta \right|$$ + C

 

[cristo]

If you use the substitution I suggested, then you end up with $$\int\frac{dy}{\sqrt{1+y^2}}$$. This is a standard integral, is it not?

 

[Brickster]

If you use the substitution I suggested, then you end up with $$\int\frac{dy}{\sqrt{1+y^2}}$$. This is a standard integral, is it not?

Yes, I'm fairly certain that is a standard integral, and it comes out to:

$$sin^{-1}y + C$$

But I'm also fairly certain that I am supposed to solve this problem using the expression, substitution, and identity when I do the trig sub. Maybe I am overcomplicated the problem and should just stick to the arcsin answer, Gib Z's method seems to be the one I'm after

 

[cristo]

Yes, I'm fairly certain that is a standard integral, and it comes out to:

$$sin^{-1}\frac{y} + C$$

Close; the solution is $$\sinh^{-1} y +C$$

 

[Brickster]

Close; the solution is $$\sinh^{-1} y +C$$

My mistake, you're right. Thanks!

 

[Gib Z]

It's always nice to know alternative methods, because we can know express the inverse hyperbolic sine function in terms of the natural logarithm =] Which we could have also done more directly, but o well!