Author Topic: Intergration + algebra (Read 998 times) Tweet Share

Chavi · « **on:** May 16, 2010, 05:55:43 pm »

Can I please get some help with the following questions:
1. Differentiate $f(x) = sin(x)cos^{n-1}(x)$

Hence evaluate: $\int_{0}^{\frac{\pi}{2}}sec^4(x)\, dx$

2) Let $U_n = \int_{0}^{\frac{\pi}{4}}tan^2(x)\, dx$ where $n\epsilon Z, n>1$
a) Express $U_n + U_{n-2}$ in terms of n
b) Hence show that $U_6 = \frac{13}{5} - \frac{\pi}{4}$

Thanks in advance!

kenhung123 · « **Reply #1 on:** May 16, 2010, 06:25:48 pm »

That looks hardcore..

Martoman · « **Reply #2 on:** May 16, 2010, 06:51:18 pm »

Eh it is madly, madly hardcore.

Ok so we start with $f'(x) = cos(x)cos^{n-1}(x) + sin(x)(-sin(x)(cos^{n-1-1}(x))(n-1)$
Via product/chain rule. Trying to get it all in terms of cos's.

$f'(x) = cos^n{x} -sin^{2}cos^{n-2}(n-1)$
$= cos^n(x) - (1-cos^2(x))(cos^{n-2}(x))(n-1)$
$=cos^n(x) -(cos^{n-2}(x) - cos^n(x}))(n-1)$
$= cos^n(x) + cos^n(x)(n-1) - (n-1)(cos^{n-2}(x)$
$= cos^n(x)(1+n-1) - (n-1)cos^{n-2}(x)$
$= ncos^n(x) - (n-1)cos^{n-2}(x)$

Integrate all sides.

$sin(x)cos^{n-1}(x) = n\int cos^n(x)dx - (n-1) \int cos^{n-2}(x)dx$
SO finally $sin(x)cos^{n-1}(x) - n \int cos^n(x)dx = -(n-1) \int cos^{n-2}(x)dx$

I put it in this form because after playing with it, its easier to let n = -2.

Now:

$[sin(x)cos^{-3}(x)]^{\frac{\pi}{4}} _{0} +2\int_{0}^{\frac{\pi}{4}} sec^2(x)dx = 3\int_{0}^{\frac{\pi}{4}} sec^4(x)dx$

When crunched gets $2 + 2 = 3\int_{0}^{\frac{\pi}{4}} sec^4(x)dx$

FINALLY $\int_{0}^{\frac{\pi}{4}} sec^4(x)dx = \frac{4}{3}$

IF THIS IS WRONG I WILL HARM SOMEBODY IN MY NEAR VICINITY!!!! (working on question 2 now)

edit: Is it meant to be $tan^n(x)$ ? otherwise, I don't know how to do n-2 without an n in there somewhere

Cthulhu · « **Reply #3 on:** May 16, 2010, 07:03:38 pm »

Quote from: Martoman on May 16, 2010, 06:51:18 pm

edit: Is it meant to be $tan^n(x)$ ? otherwise, I don't know how to do n-2 without an n in there somewhere

This has me stumped too.

Martoman · « **Reply #4 on:** May 16, 2010, 07:05:25 pm »

Yeah i'm certain it has to be otherwise wtf?

Cthulhu · « **Reply #5 on:** May 16, 2010, 07:17:31 pm »

Unless its $\tan^{2n}(x)$

Martoman · « **Reply #6 on:** May 16, 2010, 07:19:20 pm »

Quote from: Martoman on May 16, 2010, 07:05:25 pm

Yeah i'm certain it has to be otherwise wtf?

Ok asuming that is the case it becomes quite elegant.

$U_n +U_{n-2} = \int_{0}^{\frac{\pi}{4}} tan^n(x) +tan^{n-2}(x)$

In factorising, take out the lowest power, as always.

$=\int_{0}^{\frac{\pi}{4}} tan^{n-2}(x)(tan^2(x)+1)$
$= \int_{0}^{\frac{\pi}{4}} tan^{n-2}(sec^2(x))$

Awesome substitution tells us that $u = tan(x)$ so $du = sec^2(x) dx$

ooooo so $\int_{0}^{1} u^{n-2}du$

= $\frac{1}{n-1}$

Second part requires a bit of creativity, i'll leave this for you to do unless you really want me to.

Hint $U_n +U_{n-2} = \frac{1}{n-1}$ as we just worked out. This seems obvious but you need to be aware of this fully to work out the second part. And it does in fact work out to what you say.

So DEFINATELY there i a typo there. It has to be $tan^n(x)$ not $tan^2(x)$

Chavi · « **Reply #7 on:** May 16, 2010, 10:14:39 pm »

Thanks Martoman. That was def a typo with tan.
Now How The F do you add Karma??
lol

TrueTears · « **Reply #8 on:** May 16, 2010, 10:29:43 pm »

u need 50 posts

theuncle · « **Reply #9 on:** May 17, 2010, 08:41:12 pm »

The original differentiation can be made simpler by doing some rearranging
$f(x)=sin(x)cos^{n-1}(x)$
$=\frac{sin(x)cos^n(x)}{cos(x)}$
$=tan(x)cos^n(x)$

Chavi · « **Reply #10 on:** May 17, 2010, 09:55:43 pm »

Quote from: TrueTears on May 16, 2010, 10:29:43 pm

u need 50 posts

i meant to give martoman karma

Martoman · « **Reply #11 on:** May 17, 2010, 10:41:47 pm »

one more...

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