Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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!

share|improve this question
3  
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
1  
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
add comment

2 Answers

up vote 3 down vote accepted

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

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

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

share|improve this answer
1  
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
1  
@Kudla69 see my edit. –  user2468 Jul 26 '12 at 2:58
2  
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
show 3 more comments

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.