# evaluate the integral: $\int{(3\csc(x)\cot(x) - 5x^7 +\frac{4}{x} + 3)dx}$

$$\int{(3\csc(x)\cot(x) - 5x^7 +\frac{4}{x} + 3)dx}$$

I know this is a simple problem, but I don't have the answer for it and I just want to make sure that I'm correct!

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Yes, it is a simple problem, at least if you replace $dy$ by $dx$. And it is even simpler if you don't, because one then gets to assume the $x$ stuff is a constant. –  André Nicolas Jul 26 '12 at 2:16
@AndréNicolas Really, its easier if it is $dy$! However, it does look like a mistake. –  Argon Jul 26 '12 at 2:18
If you just want to verify your answers, use this: integrals.wolfram.com/index.jsp –  Vectk Jul 26 '12 at 2:23
Since sometimes equivalent answers can look different, it's often better to verify by differentiating the answer and checking that the result is equivalent to the original function. –  Robert Israel Jul 26 '12 at 3:05

$$\int{(3\csc(x)\cot(x) - 5x^7 +\frac{4}{x} + 3)dy} = y(3\csc(x)\cot(x) - 5x^7 +\frac{4}{x} + 3) + \text{const}.$$

So I'm assuming your integral is actually:

$$\int (3\csc(x)\cot(x) - 5x^7 +\frac{4}{x} + 3) \color{red}{dx}$$

By trig identities $$\csc(x) \cot(x) = \frac{\cos(x)}{\sin^2(x)}.$$ Now if we use $u = \sin(x)$ then $du = \cos(x)dx,$ and $\sin^2(x) =u^2.$ So $$\int \frac{\cos(x)}{\sin^2(x)} dx = \int \frac{1}{u^2} du .$$

Can you take it from here?

Edit: the complete integral is then

$$\int{(3\csc(x)\cot(x) - 5x^7 +\frac{4}{x} + 3)\ \text{d}x} = -\csc(x) - \frac{5}{8}x^8 + 4 \log(|x|) + 3x + \text{const}.$$

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can't you just do $\int{csc(x)cot(x)dx} = -csc(x)$? –  Kudla69 Jul 26 '12 at 2:47
Yes. My answer shows that $$\int \csc(x)\cot(x)dx = \int \frac{1}{u^2} du = -\frac{1}{u} = - \frac{1}{\sin(x)} = -\csc(x)$$ up to addition of a constant. –  user2468 Jul 26 '12 at 2:49
oh, well I knew that ∫csc(x)cot(x)dx=−csc(x). I was just looking for the answer of the integral so I could verify it with the one I have on my paper. –  Kudla69 Jul 26 '12 at 2:54
@Kudla69 see my edit. –  user2468 Jul 26 '12 at 2:58
If verifying your answer is all you wanted to accomplish, perhaps typing your function in wolfram alpha or similar would have been a better option –  user979616 Jul 26 '12 at 3:17

$$\forall -1\neq n\in\Bbb R\,\,,\,\int x^n\,dx=\frac{x^{n+1}}{n+1}+K\,\,,\,\int x^{-1}\,dx=\log|x|+K$$

$$\int\csc x\cot x\,dx=\int\frac{\cos x\,dx}{\sin^2 x}=\int\frac{d(\sin x)}{(\sin x)^2}=-\frac{1}{\sin x}+K\,$$

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