Hi,(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]

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

[/tex]

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

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

Then I did a substitution, [tex]y = \sin x \Leftrightarrow dy = \cos xdx[/tex] to get:

[tex]

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

[/tex]

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

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# Goniometric Integral Problem

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