The 2013 STEP Prep Thread!

That's right. Now integrate those.

That's what the was for

-Sec^2(2x)/2 + x - ??? No idea

Substitution?

That's what the was for

-Sec^2(2x)/2 + x - ??? No idea

Substitution?

That's not the integral of tan(x). Substitution would work for sec(x) but it's quite long. Try to find a similar trick as before, though it might be difficult if you're not too familiar with integrating/differentiating trig.

Anyway, I need to leave for school now so if you're still stuck when I get back I can give you more clues.

Don't you know the integral of sec x?

Is this still about the trig integral I posted?! If so, you guys have really complicated things.

Yes... It's the only.way I can do it... Because I don't have the superior knowledge you guys possess...

LOL there is no additional knowledge needed for that question. cos and sin span the same values between 0 and pi/2 (just in "reverse order), so the areas under sin/(cos+sin) and cos/(cos+sin) are exactly the same. Therefore the integral of one of them is half the integral of their sum, which is 1:

$\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{\sin x}{\cos x+\sin x}\;dx=\dfrac{1}{2}\int_0^{\frac{\pi}{2}}\;dx=\dfrac{\pi}{4}$

Nice to see that this thread is actually doing something useful now...

Probably because I haven't posted recently...

Oh wow... I wouldn't think to do that... So how do you integrate.

$\displaystyle\int_{0}^{ \pi}{2} \tan(2x) + 1 - \sec(2x) dx$

Nice to see that this thread is actually doing something useful now...

Probably because I haven't posted recently...

Without prior knowledge of the integrals of tan and sec?

$\bullet\;\;\displaystyle\int \tan 2x\;dx=\int \dfrac{\sin 2x}{\cos 2x}\;dx=-\dfrac{1}{2}\ln \cos 2x$


$\bullet\;\;\displaystyle\int \sec 2x\;dx=\dfrac{1}{2}\int \dfrac{\sec 2x(\sec 2x +\tan 2x)}{\sec 2x+\tan 2x}\;dx$

$\;\;\;\;\displaystyle=\dfrac{1}{2}\int \dfrac{\sec^2 2x+\sec 2x\tan 2x}{\sec 2x+\tan 2x}\;dx=\dfrac{1}{2}\ln (\sec 2x+\tan 2x)$

how would I know that? Do I learn it?

Without prior knowledge of the integrals of tan and sec?

$\bullet\;\;\displaystyle\int \tan 2x\;dx=\int \dfrac{\sin 2x}{\cos 2x}\;dx=-\dfrac{1}{2}\ln \cos 2x$


$\bullet\;\;\displaystyle\int \sec 2x\;dx=\int \dfrac{\sec 2x(\sec 2x +\tan 2x)}{\sec 2x+\tan 2x}\;dx$

$\;\;\;\;\displaystyle=\int \dfrac{\sec^2 2x+\sec 2x\tan 2x}{\sec 2x+\tan 2x}\;dx=\dfrac{1}{2}\ln (\sec 2x+\tan 2x)$

Typo, I corrected whilst you were quoting me I think

You don't need to learn it at all - it's not that obscure. When you have an integral it's useful to think: "what would I need to get it into this form" (i.e. into a standard derivative form: chain, product, log, depending on the integral). From there, it is quite straightforward to see what might work.

Now you're back, please give us some useless stuff to munch on