What is the best approach for solving the goniometric integral problem?

[TD]

Hi,

I'm having some trouble with solving this indefinite integral.

$$\int {\sqrt {\frac{{6\cos ^2 x + \sin x\cos (2x) + \sin x}}{{2 - \sin x}}} } dx$$

I was able to lose the sin(x) and get a cos(x) out of the square root by doing this:

$$\int {\sqrt {\frac{{6\cos ^2 x + \sin x(2\cos^2 x -1) + \sin x}}{{2 - \sin x}}} } dx$$

$$\int {\sqrt {\frac{{6\cos ^2 x + 2\sin x\cos^2 x -\sin x + \sin x}}{{2 - \sin x}}} } dx$$

$$\int {\sqrt {\frac{{\cos ^2 x(6 + 2\sin x)}}{{2 - \sin x}}} } dx$$

$$\int \cos x{\sqrt {\frac{{6 + 2\sin x}}{{2 - \sin x}}} } dx$$

Then I did a substitution, $$y = \sin x \Leftrightarrow dy = \cos xdx$$ to get:

$$\int {\sqrt {\frac{{6 + 2y}}{{2 - y}}} } dy$$

I think I can say this looks a lot better than the initial integral, but after that I used about 3 more substitutions and 2 sides of paper. I finally got something but it wasn't all correct I'm affraid.
I was wondering if I started out wrong or if someone sees an easy way to continue (or an easier to start).

Naughty integral if you ask me

 

[dextercioby]

It looks okay so far.Now try the sub

$$\frac{6+2y}{2-y}=t^{2}$$

Daniel.

 

[TD]

Hi, perhaps I should've went on but that's indeed what I did first.

I then got:

$$\int {\frac{{20t^2}}{{(t^2+2)^2}} } dt$$

I was able to expand that to:

$$\int {\frac{{20}}{{t^2+2}} - \frac{{40}}{{(t^2+2)^2}}} dt$$

The first one is no problem, I can easily get there with a ArcTan like this:

$$\int {\frac{{20}}{{t^2+2}}} dt = 10\sqrt 2 \arctan \left( {\frac{{\sqrt 2 t}}{2}} \right) (+ C)$$

But what about the second one? I know it gived an ArcTan as well, but not only that. I've seen recursion formula's for it but is there a way to do it 'by hand'?

PS: I read you're temporarily in Leuven, I study in Brussels - that's very nearby

 

[dextercioby]

U could have easily done it using part integration

$$10\int t \ \frac{d\left(t^{2}+2\right)}{\left(t^{2}+2\right)^{2}} =-10\frac{t}{t^{2}+2} +10\int \frac{dt}{t^{2}+2}$$

Can you take it from here...?

Daniel.

 

[TD]

Of course! Then I get:

$$\int {\frac{{20t^2}}{{(t^2+2)^2}} } dt = 5\sqrt 2 \arctan \left( {\frac{{\sqrt 2 t}}{2}} \right) - \frac{{10t}}{{t^2 + 2}} + C$$

Susbstituting back:

$$\frac{6+2y}{2-y}=t^{2} \Leftrightarrow t=\sqrt{\frac{6+2y}{2-y}} \,\,\, gives:$$

$$5\sqrt 2 \arctan \left( {\sqrt {\frac{{y + 3}}{{2 - y}}} } \right) + \sqrt 2 \left( {y - 2} \right)\sqrt {\frac{{y + 3}}{{2 - y}}} + C$$

Susbstituting back:

$$y=\sin x \,\,\, gives:$$

$$5\sqrt 2 \arctan \left( {\sqrt {\frac{{\sin x + 3}}{{2 - \sin x}}} } \right) + \sqrt 2 \left( {\sin x - 2} \right)\sqrt {\frac{{\sin x + 3}}{{2 - \sin x}}} + C$$

I hope that's it. Finally, thanks

 

[Zurtex]

Yep, took a bit of messing to check it was the right, but it is.

 

[TD]

Yes, I had Mathematica take the derivative and after FullSimplify, the initial expression almost identically rolled out. Just had to put the cos (and a factor) back in the square root