Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2) Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)} = -(1/2)ln(1-u^2) + C = -(1/2)ln(sin^[2]x) +C = ln(cosec x) +C Yet differentiating back gives -cot x. Why does this substitution not work?
First: [tex] csc x = \frac{1}{sin x} [/tex] NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this: [tex] u = sin x dx [/tex] [tex] du = cos x dx [/tex] [tex] sec x du = dx [/tex] [tex] \frac{1}{\sqrt{1-u^2}} du = dx [/tex] So this makes the integral: [tex] \int \frac{1}{u\sqrt{1-u^2}} du [/tex] See if you can take it from there.
My query was concerning the substitution u = cos x , rather than u = sin x . I did make a mistake, arcos x = -1/sqrt(1-x^2). I've edited my original post.
well your subsitution is didn't work because it is difficult to solve integration with assumption not being in the question
Because you make an equality out of the following non-equality: [tex]-\int\frac{du}{1-u^{2}}\neq\frac{-1}{2}ln(1-u^{2})+C[/tex]
Help please I can't :(. I've spent most of this afternoon trying this question and some other one. I know I could just use the set result, but I want to understand it, and I don't see how to integrate that last result at all :(.
There is a special method to this. [itex]\int \mathrm{cosec} x \ \mathrm{d}x[/itex] If you multiply this by [itex]\frac {\mathrm{cosec} x - \cot x}{\mathrm{cosec} x - \cot x}[/itex] and simplify the numerator you will get an integral of... [itex]\int \frac{\mathrm{cosec} ^2 x - \mathrm{cosec} x \cot x}{\mathrm{cosec} x - \cot x} \ \mathrm{d}x[/itex] Substitute [itex]u = \mathrm{cosec} x - \cot x[/itex] and it should work out beautifully. Carry on from here and post back if you still need help.
Thanks :)! Thanks, that worked out really well :). I was wondering though, how would one work it out from the form [tex] \int \frac{1}{u\sqrt{1-u^2}} du [/tex] ? Could anyone help me see how to integrate it from this?