Something my teacher can't explain to me

Hi, I'm not sure I'm very good at wording this question, as I've tried to ask it before and have not been understood. I'll stiill give it a shot, as it is frustrating that I don't know the answer; <.

When I was taught how to differentiate, was I was told to multiply by the value of the indice, then reduce the power by 1, for example.
$\frac{d}{dx} x^2 = 2x$

Then, when I was taught to intergrate, I was told to increase the power by 1, and then divide by the new power, for example
$\displaystyle\int 2x\ dx = x^2$

Then when I was taught to differentiate trigonometric functions like sin squared x, I was told to multiply by the power, reduce the power by 1, and multiply by the differential of whats "in the bracket", for example,

$\frac{d}{dx} sin^2 x\ dx = 2 sinx cosx$

However, when I integrate $sin^2 x$
I don't increase the power by 1, then divide by the new power, then divide by the differential for whats "in the bracket," for example:

$\displaystyle\int sin^2 x\ dx = sin^3x/3cosx = tanx sin^2 x /3$

Why is this? In my FP2 module I've been taught ways of integrating trigonometric functions with powers etc, but why is that way that worked for everything else suddenly incorrect?

Because the rules you have learnt are only applied to very specific cases.

Oh and Farhan, if you actually read his post he isn't asking how... just why it doesn't work.

You've gotten to FP2 without understanding the chain rule from C3?

I'm not too sure why you're considering the integral of $\sin ^2x$ after considering it's derivative, the two results are not related.

To integrate $sin ^2x$, you need to note that $\cos 2x = 1-2\sin ^2x$ and rearrange for $sin ^2x$. Then try to integrate that.

The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand. The more formal way of writing this is:

$\displaystyle\int f'(g(x))g'(x)dx = f(g(x))+C$

Well I actually studied FP2 and C3 simultaneously, and I've often studied modules "the wrong way around," meaning I've often learned results before I've been able to derive them, which has confused me a bit. But as far as I know, I understand the chain rule from c3, but if I went the long way about differentiating sin^2 x I'd do it like this:

$y=sin^2 x[br]let u=sinx[br]y=u^2$
$\frac{dy}{du} = 2u$
$\frac{du}{dx} = cosx$
$\frac{dy}{dx} = 2sinx cosx$

but what I was trying to say in my first post is that this rule of thumb thing where you differentiate $(sinx)^2$ by multiplying by the power then reduce the power by 1, and multiply by the differential of the bracket seems to work (I have multiplied by a function of x here), yet it doesn't work for integration, the other way around.

I can definitley see why it does work for differentiation, it's just a shortened version of the chain rule. You say "The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand," but you're happy to do this the other way around with differentiation. Why is that?

Also thanks for the help so far everyone =)

Hi, I'm not sure I'm very good at wording this question, as I've tried to ask it before and have not been understood. I'll stiill give it a shot, as it is frustrating that I don't know the answer; <.

When I was taught how to differentiate, was I was told to multiply by the value of the indice, then reduce the power by 1, for example.
$\frac{d}{dx} x^2 = 2x$

Then, when I was taught to intergrate, I was told to increase the power by 1, and then divide by the new power, for example
$\displaystyle\int 2x\ dx = x^2$

Then when I was taught to differentiate trigonometric functions like sin squared x, I was told to multiply by the power, reduce the power by 1, and multiply by the differential of whats "in the bracket", for example,

$\frac{d}{dx} sin^2 x\ dx = 2 sinx cosx$

However, when I integrate $sin^2 x$
I don't increase the power by 1, then divide by the new power, then divide by the differential for whats "in the bracket," for example:

$\displaystyle\int sin^2 x\ dx = sin^3x/3cosx = tanx sin^2 x /3$

Why is this? In my FP2 module I've been taught ways of integrating trigonometric functions with powers etc, but why is that way that worked for everything else suddenly incorrect?

Well I actually studied FP2 and C3 simultaneously, and I've often studied modules "the wrong way around," meaning I've often learned results before I've been able to derive them, which has confused me a bit. But as far as I know, I understand the chain rule from c3, but if I went the long way about differentiating sin^2 x I'd do it like this:

$y=sin^2 x[br]let u=sinx[br]y=u^2$
$\frac{dy}{du} = 2u$
$\frac{du}{dx} = cosx$
$\frac{dy}{dx} = 2sinx cosx$

but what I was trying to say in my first post is that this rule of thumb thing where you differentiate $(sinx)^2$ by multiplying by the power then reduce the power by 1, and multiply by the differential of the bracket seems to work (I have multiplied by a function of x here), yet it doesn't work for integration, the other way around.

I can definitley see why it does work for differentiation, it's just a shortened version of the chain rule. You say "The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand," but you're happy to do this the other way around with differentiation. Why is that?

Also thanks for the help so far everyone =)