CatLove9
Integrate. sqrt(1+x^2)/x



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CatLove9
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I think it should be \[\int\limits_{}^{} 1/\sin(x)\cos^2x\]

Mimi_x3
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Or start from \[\int\frac{\sqrt{1+x^{2}}}{x}
dx\]

CatLove9
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You could started from there, I got up to 1/sinxcos^2x

Mimi_x3
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Assuming that you did the other part correct then..
\[\int \frac{1}{sinxcos^2{x}}\]
wait..are you sure about that though?

lgbasallote
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dw:1342236036378:dw
\[\sec \theta = \sqrt{1+ x^2}\]
\[\tan \theta = x\]
\[\sec^2 \theta d\theta = dx\]
\[\int \frac{\sqrt{1+x^2}}{x}dx \implies \frac{\sec \theta(\sec^2 \theta d\theta )}{\tan \theta}\]

CatLove9
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I'm sure that \[\int\limits_{}^{} \sec^3x/\tan(x)\]

CatLove9
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That turns into 1/cos^2xsinx

lgbasallote
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it does?

lgbasallote
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i mean yeah it does

CatLove9
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I know I can't do u sub.

lgbasallote
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there's only one thing you can do...integration by parts

lgbasallote
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\[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \csc x \sec^2 x dx\]

CatLove9
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But can I integrate that?

lgbasallote
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well i was just spitting out ideas....

CatLove9
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That's why I didn't change it to that because I knew there was nothing to do there.

lgbasallote
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well i know i can make 1 = sin^2 x + cos^2 x

lgbasallote
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\[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \frac{ \sin^2 x + \cos^2 x}{\sin x \cos^2 x}dx\]

CatLove9
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OH! That's good

lgbasallote
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\[\int \frac{\sin^2 x}{\sin x \cos ^2 x}dx + \int \frac{\cos^2 x}{\sin x \cos^2 x}dx\]

lgbasallote
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yup saw the answer now :D i'll let you do the good stuff ;)

CatLove9
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Ah, yes, yes, I see it too.

lgbasallote
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nice! congrats