Strategies for Solving Integrals with Trigonometric Functions

[Mentallic]

Homework Statement

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

The Attempt at a Solution

This problem is part of a bigger picture, and I can't seem to figure out how to approach this integral.

 

[Dickfore]

It's expressible in terms of the incomplete elliptic integral of the second kind:

$$E(\phi, k) = \int_{0}^{\phi}{\sqrt{1 - k^{2} \, \sin^{2} t} \, dt}$$

Hint:

Express the cosine squared in terms of sine squared and then divide by the free term under the squared root to

 

[Mentallic]

In other words, not expressible in terms of elementary functions. Looks like my calculator is of no use then, gay...

So the answer is $$\sqrt{2}E\left(1,\frac{1}{2}\right)$$

How could I go about finding an approximation for this?

 

[Feldoh]

Here's the approximation according to Mathematica to 50 decimal places:

1.3114424982155470455454946537619651179489905076619

N[Sqrt[2] EllipticE[1, 1/2], 50]

is the command used...

 

[Dickfore]

In other words, not expressible in terms of elementary functions. Looks like my calculator is of no use then, gay...

So the answer is $$\sqrt{2}E\left(1,\frac{1}{2}\right)$$

How could I go about finding an approximation for this?

I think you made a mistake. FIrst of all, your $k$ is wrong. Secondly, I don't know how you found that upper limit, since you had an indefinite integral.

 

[Mentallic]

I think you made a mistake. FIrst of all, your $k$ is wrong. Secondly, I don't know how you found that upper limit, since you had an indefinite integral.

Nope, I'm fairly certain my k is correct and I originally posted the indefinite integral assuming I wouldn't need help with evaluating the limits, they were 0 to 1 as you'd expect.

 

[Dickfore]

Ok then, cool. Have a nice life.