tricky integral

**mathcore** · February 21st 2011, 02:35 PM

what do i do here?

S 1/2sin^2(x) dx

the S is meant to be the integrate zigzag thing so its integrate 1/2sin^2(x) dx? i am lost. i calculated the answer to be 1/2tan(x) using software but i don't get why so can someone talk me through a step by step?

**pickslides** · February 21st 2011, 02:45 PM

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

**mathcore** · February 21st 2011, 03:15 PM

but it is 1 / (2sin^2(x)) so what do i get?

1 / 2((1 - cos2x)/2)

cancels to 1/1 - cos2x). i dont still know what to do when you integrate something in the bottom unless its exact derivative on top then you use logs but if it isn't... :s

is it a substitution?

thanks so far

**e^(i*pi)** · February 21st 2011, 03:39 PM

Yes, let $u = 1-\cos(2x)$

**topsquark** · February 21st 2011, 03:42 PM

Originally Posted by mathcore

but it is 1 / (2sin^2(x)) so what do i get?

1 / 2((1 - cos2x)/2)

cancels to 1/1 - cos2x). i dont still know what to do when you integrate something in the bottom unless its exact derivative on top then you use logs but if it isn't... :s

is it a substitution?

thanks so far

So the problem is actually
$\displaystyle \int \frac{1}{sin^2(x)}dx$

(Please make better use of parethesis!)

You might recognize that
$\displaystyle \int \frac{1}{sin^2(x)}dx = \int csc^2(x) dx = -cot(x) + C$

If not, then
$\displaystyle \int \frac{1}{sin^2(x)}dx = \int \frac{sin^2(x) + cos^2(x)}{sin^2(x)}dx$

Now let y = cos(x) and see what happens.

-Dan

(This is probably the longer way around.)

**e^(i*pi)** · February 21st 2011, 03:48 PM

Take a factor of 1/2 out the integral then continue as topsquark did above

$\displaystyle \int \dfrac{1}{2 \sin^2(x)} = \int \left( \dfrac{1}{2} \cdot \csc^2(x)\right) = \dfrac{1}{2} \int \csc^2(x)$

You also have the option of using the double angle identity and a sub as per posts 2 and 4

edit: corrected "sec" to "csc"

**mathcore** · February 21st 2011, 03:56 PM

Originally Posted by e^(i*pi)

Take a factor of 1/2 out the integral then continue as topsquark did above

$\displaystyle \int \dfrac{1}{2 \sin^2(x)} = \int \left( \dfrac{1}{2} \cdot \sec^2(x)\right) = \dfrac{1}{2} \int \sec^2(x)$

You also have the option of using the double angle identity and a sub as per posts 2 and 4

if you mean csc instead of sec then the solution i have works... can you tell me is it possible you meant csc?

my solution is -1/(2tan(x)).

**e^(i*pi)** · February 21st 2011, 03:58 PM

Originally Posted by mathcore

if you mean csc instead of sec then the solution i have works... can you tell me is it possible you meant csc?

my solution is -1/(2tan(x)).

Yes it is csc, such is what happens when doing calculus at 1am. Thanks for pointing it out.

Your answer is almost correct - what do you need to add for indefinite integrals?

**mathcore** · February 21st 2011, 04:04 PM

no i dont need a constant in this case because its part of a seperating variables d.e., i have the constant on the other side

**topsquark** · February 21st 2011, 04:41 PM

Originally Posted by mathcore

what do i do here?

S 1/2sin^2(x) dx

Originally Posted by mathcore

no, the problem is NOT 1/sin^2(x), it is 1/(2sin^2(x)) which is what i Said. do NOT call into question my parenthesis when you havent got a clue. my parenthesis was correct.

Funny how that 1/2sin^2(x) is $\displaystyle \frac{1}{2} sin^2(x)$ and 1/(2sin^2(x)) is $\displaystyle \frac{1}{2 sin^2(x)}$ . These are obviously not the same. Speaking of obvious
$\displaystyle \int \frac{1}{2 sin^2(x)} dx = \frac{1}{2} \int \frac{1}{sin^2(x)} dx$

I expected you to be able to recognize that I factored the 2 out and worked on the meat of the problem. Evidently I expected too much from you.

-Dan

**Chris L T521** · February 21st 2011, 05:12 PM

The question has been answered. This is yet another good reason why members should learn the basics of LaTeX, so that no miscommunication occurs.

Thread Closed.

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