Trigonometric integral question

• Aug 23rd 2010, 03:23 PM
Solid8Snake
Trigonometric integral question
Greetings, I have come across a question that I for some reason cannot solve. The question states to find the integral of (csc(x))/(csc(x))-sin(x)) dx. I know the answer is tangent but I have no clue what trig identities were used to come across to that answer, any help would be greatly appreciated.

• Aug 23rd 2010, 03:46 PM
skeeter
Quote:

Originally Posted by Solid8Snake
Greetings, I have come across a question that I for some reason cannot solve. The question states to find the integral of (csc(x))/(csc(x))-sin(x)) dx. I know the answer is tangent but I have no clue what trig identities were used to come across to that answer, any help would be greatly appreciated.

$\displaystyle \frac{\sin{x}}{\sin{x}} \cdot \frac{\csc{x}}{\csc{x}-\sin{x}} =$

$\displaystyle \frac{1}{1 - \sin^2{x}} =$

$\displaystyle \frac{1}{\cos^2{x}} =$

$\sec^2{x}
$

able to finish?
• Aug 23rd 2010, 03:53 PM
Solid8Snake
thanks so much, I can't believe I couldn't solve it myself. But I had another question, I was just wondering if we always have to put the integral sign aswell as dx aslong as we are changing the way the problem looks for example by the use of trig identities. Do we always have to write the integral sign as long as we are solving?

Thanks again.
• Aug 23rd 2010, 03:57 PM
Prove It
If you are simplifying the integrand, you don't need the integral signs and the $dx$ as long as your equalities are correct.

Once you have simplified it though, then you'd have to make a statement about the integrals being equal.

So in this case, after you've found $\frac{\csc{x}}{\csc{x} - \sin{x}} = \sec^2{x}$

then you would say

$\int{\frac{\csc{x}}{\csc{x} - \sin{x}}\,dx} = \int{\sec^2{x}\,dx}$.
• Aug 23rd 2010, 04:56 PM
Quote:

Originally Posted by Solid8Snake
Greetings, I have come across a question that I for some reason cannot solve. The question states to find the integral of (csc(x))/(csc(x))-sin(x)) dx. I know the answer is tangent but I have no clue what trig identities were used to come across to that answer, any help would be greatly appreciated.

$\displaystyle\int\frac{Cosec(x)}{Cosec(x)-Sin(x)}dx$
$\displaystyle\huge\frac{\frac{1}{Sin(x)}}{\frac{1} {Sin(x)}-Sin(x)}=\frac{\frac{1}{Sin(x)}}{\frac{1}{Sin(x)}-\frac{Sin^2(x)}{Sin(x)}}$
$=\displaystyle\huge\frac{\left(\frac{1}{Sin(x)}\ri ght)}{\left(\frac{1}{Sin(x)}\right)}\ \frac{1}{1-Sin^2(x)}=\frac{1}{Cos^2(x)}=Sec^2(x)$