Hi Guys,

Sorry to bother you again. Please look at the question below. I only managed to find part (i). Please advice me on how to find part ii and iii. Thanks for your help

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- Nov 19th 2007, 08:16 PMaeonboyPlease help... challenging Integral Question
Hi Guys,

Sorry to bother you again. Please look at the question below. I only managed to find part (i). Please advice me on how to find part ii and iii. Thanks for your help - Nov 19th 2007, 09:39 PMJhevon
to part (ii), just find the integral and see what you get

$\displaystyle \int \sin^3 x ~dx = \int \sin^2 x \sin x ~dx$

$\displaystyle = \int \left( 1 - \cos^2 x \right) \sin x~dx$

can you continue?

for part (iii). notice that the sly devils let you calculate the derivative of one of the functions in the product and the integral of the other function in the product. this is exactly what you would do for integration by parts. so that is what you want to use here.

recall, $\displaystyle \int uv' = uv - \int u'v$

where $\displaystyle u$ and $\displaystyle v$ are functions - Nov 19th 2007, 09:44 PMaeonboy
alright , i will try to do it now.. thanks

- Nov 19th 2007, 11:05 PMaeonboy
Sorry, i still cant do it for part ii.. can anyone pls continue the steps?

thanks alot - Nov 19th 2007, 11:14 PMCaptainBlack
- Nov 19th 2007, 11:17 PMaeonboy
Sorry, i still don't know how to set it.. could u pls give more advice? thanks

- Nov 20th 2007, 06:42 AMaeonboy
can anyone help me with part ii and iii. I do not know how to do it.. thanks

- Nov 20th 2007, 08:40 AMJhevon
at least tell us what you've tried before. if need be, i'll give you the solution. we just want to know that you're not just looking for someone to do your homework, but that you're actually trying. you should be able to at least do part (ii), which just requires a substitution of $\displaystyle u = \cos x$.

for part (iii), i gave you the formula. here, $\displaystyle u = \ln (\sec x + \tan x)$ and $\displaystyle v' = \sin^3 x$. part (i) asked you to find $\displaystyle u'$ and part (ii) asked you to find $\displaystyle v$. just plug them into the formula i gave you - Nov 20th 2007, 09:58 AMCaptainBlack

$\displaystyle \frac{d}{dx} \left[p \cos^3(x) + q \cos^2(x) + r \cos(x)+s\right]$$\displaystyle

=-3p\cos^2(x) \sin(x) - 2 q \cos(x) \sin(x) - r \sin(x)

$

............. $\displaystyle

=-3p(1-\sin^2(x))\sin(x)- 2 q \cos(x) \sin(x) - r \sin(x)

$

............. $\displaystyle

=-3p\sin(x)+3p\sin^3(x)- 2 q \cos(x) \sin(x) - r \sin(x)

$

So if this is equal to $\displaystyle \sin^3(x)$ then $\displaystyle p=1/3,\ r=-1,\ q=0 $.

Easy eh?

RonL - Nov 20th 2007, 04:20 PMaeonboy
Thanks CaptainBlack for the help. I did try before and i do not know how to do it.

Is there any other ways to solve this problem? Could we use this way.

$\displaystyle

\int \sin^3 x ~dx = \int \sin^2 x \sin x ~dx

$

$\displaystyle

= \int \left( 1 - \cos^2 x \right) \sin x~dx

$

Can anyone guide me how to use this way to solve this problem ?

Thanks - Nov 20th 2007, 06:06 PMJhevon
substitution.

let $\displaystyle u = \cos x$

$\displaystyle \Rightarrow du = - \sin x ~dx$

$\displaystyle \Rightarrow -du = \sin x ~dx$

so our integral becomes:

$\displaystyle - \int \left( 1 - u^2 \right)~du$

$\displaystyle = \int \left( u^2 - 1 \right)~du$

i suppose you can integrate this, it's just the power rule. when finshed, back substitute $\displaystyle \cos x$ for $\displaystyle u$