Integration by Parts: \int{\frac{xcosx}{sin^2x}dx}

[jetpac]

Homework Statement

Use integration by parts to evaluate the following integral:

$$\int{\frac{x cos x}{sin^2 x}dx}$$

Homework Equations

$$\int{udv} = uv - \int{v du}$$

The Attempt at a Solution

Select U according to the order:

L - logarithmic, a - algebraic, t - trigonometric, e - exponential.

So possible contender for u would by x, leaving dv = $$\frac{cosx}{sin^2x}$$

so du/dx= 1 => du = dx, v = $$\int{\frac{cosx}{sin^2x}}$$

= $$\int\frac{1}{sinx}*\frac{cosx}{sinx} = \int\frac{1}{sinx}*cotx$$

That's kind of where I'm lost if anyone can help, I'd really appreciate it.

Thanks!

 

[Tedjn]

The integral

$$\int \dfrac{\cos x}{\sin^2 x}\,dx$$​

can be done with a substitution. Do you see how?

 

[jetpac]

Would i be on the right track to let u = 1/sinx and dv = cotx dx so v = ln|sinx|? Maybe it sounds stupid but I can't think right now how to go about differentiating 1/sinx. Would it be something like the chain rule, so -cosx * (1/sinx) * err.. a bit lost here again sorry!

 

[Tedjn]

Ah, you are trying to integrate by parts again, but I should have been more specific -- it can be done by u-substitution. That is, you let u = something and du = something dx and substitute. If we think about this carefully, then it is true that this is argument is supported by the chain rule.

 

[Apphysicist]

Would i be on the right track to let u = 1/sinx and dv = cotx dx so v = ln|sinx|? Maybe it sounds stupid but I can't think right now how to go about differentiating 1/sinx. Would it be something like the chain rule, so -cosx * (1/sinx) * err.. a bit lost here again sorry!

So I like how in your first post, you went ahead and wrote out the integration by parts [let v = int(cos(x)/(sin(x)^2)dx)]:

x*v-int(v dx)

Clearly, from there you need to be able to do the integration in v. I also like how you started considering cot(x)...maybe if you rewrote the integrand of v using cot(x) and another function, you'll notice a familiar derivative?

After that, you need a bit of finesse to finish int(v dx)...hint: multiply by "1" in order to make a u-sub.