1. ## trig integral

$\displaystyle \int \sin^2 (5x) dx$
I think I should use the identity:
$\displaystyle \sin^2 (x) = \frac {1}{2}(1 - \cos x)$
...and rewrite the integral like this:
$\displaystyle \frac {1}{2} \int (1 - \cos (10x)) dx$
But then what I use for u to substitute? I tried using u = cos(10x) and I don't think that's right.
Can you point me in the right direction? Thank you!

2. $\displaystyle \int \sin(kx)dx=-\frac {1}{k} \cos(kx) +c$

3. I would like to start out by saying that I am completely against making u substutions for all integrals...

I would much rather recognize a pattern, guess at an antiderivative, and then put a correction factor out in front (I know most universities don't teach it like this, but let's give it a shot)

$\displaystyle \int \sin{(5x)}dx$

We know the integral of $\displaystyle \sin x$ is $\displaystyle -\cos x$

So let's "guess" that the integral is $\displaystyle -\cos (5x)$

Now if we take the derivative of that, we get $\displaystyle 5\sin (5x)$, so we're off by a factor of 5, so let's put $\displaystyle \frac{1}{5}$ in front of our "guess" and we'll be good

So the answer is $\displaystyle -\frac{1}{5}\cos(5x)$

4. Originally Posted by artvandalay11
I would like to start out by saying that I am completely against making u substutions for all integrals... Mr F asks with incredulity: ALL ......??!

I would much rather recognize a pattern, guess at an antiderivative, and then put a correction factor out in front (I know most universities don't teach it like this, but let's give it a shot)

$\displaystyle \int \sin{(5x)}dx$

We know the integral of $\displaystyle \sin x$ is $\displaystyle -\cos x$

So let's "guess" that the integral is $\displaystyle -\cos (5x)$

Now if we take the derivative of that, we get $\displaystyle 5\sin (5x)$, so we're off by a factor of 5, so let's put $\displaystyle \frac{1}{5}$ in front of our "guess" and we'll be good

So the answer is $\displaystyle -\frac{1}{5}\cos(5x)$
Using $\displaystyle \int \cos (10x) \, dx$ as your example might have been better ....

5. This is great and thank you for replying, but it doesn't help with
$\displaystyle \int \sin^2 (5x) dx$
Yes, for a simpler integral like $\displaystyle \int \sin (5x) dx$, I don't think substitution should be used either. But isn't it the simplest way to solve this one?

6. judging from what chengbin wrote I'd say it looks like the problem was changed on me, sorry for my meaningless example go ahead and delete it if you can Mr. Fantastic

7. Originally Posted by maiamorbific
This is great and thank you for replying, but it doesn't help with
$\displaystyle \int \sin^2 (5x) dx$ Mr F says: This one is not done using a u-substitution. It's done using a standard trigonometric identity which reduces it to what I'd consider a sum of standard forms. The issue raised is how one of those standard forms is calculated. Personally, I think results like the one given in post #2 should be rote learned once an understanding of where it comes from has been established. Otherwise you're going to struggle to finish things under timed exam conditions.

Yes, for a simpler integral like $\displaystyle \int \sin (5x) dx$, I don't think substitution should be used either. But isn't it the simplest way to solve this one?
..

8. Originally Posted by maiamorbific
$\displaystyle \int \sin^2 (5x) dx$
I think I should use the identity:
$\displaystyle \sin^2 (x) = \frac {1}{2}(1 - \cos x)$
...and rewrite the integral like this:
$\displaystyle \frac {1}{2} \int (1 - \cos (10x)) dx$
But then what I use for u to substitute? I tried using u = cos(10x) and I don't think that's right.
Can you point me in the right direction? Thank you!
$\displaystyle \frac {1}{2} \int (1 - \cos (10x)) dx=\frac{1}{2}\left ( \int 1 dx +\int -\cos (10x)dx\right )$

$\displaystyle =\frac{1}{2}\left (x-\frac{1}{10}\sin (10x)\right )$

$\displaystyle =\frac{x}{2}-\frac{\sin(10x)}{20}+C$ (gotta rememeber to put that constant back in at the end)