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TD

Homework Helper

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Hi,

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

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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