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

2. Originally Posted by aeonboy
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
to part (ii), just find the integral and see what you get

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

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

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

where $u$ and $v$ are functions

3. alright , i will try to do it now.. thanks

4. Sorry, i still cant do it for part ii.. can anyone pls continue the steps?

thanks alot

5. Originally Posted by aeonboy
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
for part ii differentiate:

$p \cos^3(x) + q \cos^2(x) + r \cos(x)+s$

then set this derivative equal to $\sin^3(x)$ and solve for $p$ and $q$.

RonL

6. Sorry, i still don't know how to set it.. could u pls give more advice? thanks

7. can anyone help me with part ii and iii. I do not know how to do it.. thanks

8. Originally Posted by aeonboy
can anyone help me with part ii and iii. I do not know how to do it.. thanks
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 $u = \cos x$.

for part (iii), i gave you the formula. here, $u = \ln (\sec x + \tan x)$ and $v' = \sin^3 x$. part (i) asked you to find $u'$ and part (ii) asked you to find $v$. just plug them into the formula i gave you

9. Originally Posted by CaptainBlack
for part ii differentiate:

$p \cos^3(x) + q \cos^2(x) + r \cos(x)+s$

then set this derivative equal to $\sin^3(x)$ and solve for $p$ and $q$.

RonL

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

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

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

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

Easy eh?

RonL

10. 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.

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

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

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

Thanks

11. Originally Posted by aeonboy
Thanks CaptainBlack for the help. I did try before and i do know how to do it.

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

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

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

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

Thanks
substitution.

let $u = \cos x$

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

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

so our integral becomes:

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

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

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