Integral of -64cos^3x

So I tried to do this integral and I got 16(cos^4x)/sinx
I am guessing I'm very wrong. But I don't know why I'm wrong


Posted from TSR Mobile

There's not a nice way to say this, but yeah, what did you try?
I'd recommend writing $\cos^3(x)=\cos(x)(1-\sin^2(x))$ and then proceeding by inspection or a substitution.

So I tried to do this integral and I got 16(cos^4x)/sinx
I am guessing I'm very wrong. But I don't know why I'm wrong


Posted from TSR Mobile

Differentiation is easy (squeezing toothpaste out of the tube) and integration is hard (putting the toothpaste back into the tube)

I know you've been taught that you add one to the power, divide by the new power and the divide by the derivative. But that's just plain incorrect. It only applies to integrands of the form $(ax+b)^n$. Anything else, that "rule" doesn't hold anymore and you need to be a whole lot more creative. Differentiation is easy (squeezing toothpaste out of the tube) and integration is hard (putting the toothpaste back into the tube), there are no straightforward rules for it. The above users have given a nice way to integrate this function.

I know you've been taught that you add one to the power, divide by the new power and the divide by the derivative. But that's just plain incorrect. It only applies to integrands of the form $(ax+b)^n$. Anything else, that "rule" doesn't hold anymore and you need to be a whole lot more creative. Differentiation is easy (squeezing toothpaste out of the tube) and integration is hard (putting the toothpaste back into the tube), there are no straightforward rules for it. The above users have given a nice way to integrate this function.

This is what I did...why is it wrong?

I hope you can read it


Posted from TSR Mobile

This is what I did...why is it wrong?

I hope you can read it


Posted from TSR Mobile

I know you've been taught that you add one to the power, divide by the new power and the divide by the derivative. But that's just plain incorrect. It only applies to integrands of the form $(ax+b)^n$. Anything else, that "rule" doesn't hold anymore and you need to be a whole lot more creative. Differentiation is easy (squeezing toothpaste out of the tube) and integration is hard (putting the toothpaste back into the tube), there are no straightforward rules for it. The above users have given a nice way to integrate this function.

...

Are you seriously only an a level student? You're a genius oh my god.


Posted from TSR Mobile

OP not attacking you in any way, don't worry.

But I must say I feel like C4 int is taught so badly. This chapter is so much understanding and so little method compared to a lot of the maths at A level. I've seen quite a few teachers just show their students the methods (like the rote way you've described, Zain) neglecting to give equal attention to when/(even why) each will work

Are you seriously only an a level student?

You're a genius oh my god. I have no hope

Yea it's really frustrating. I want to know why things work for a start but I think the teachers can't even explain it. Also it hardly makes them look good because if I hadn't asked you guys/tried this question I could have got something like this wrong in an exam.


Posted from TSR Mobile

By the way, you only need to use \displaystyle once in a set of $\LaTeX$ tags and it'll apply over the entire thing. :-)

OP not attacking you in any way, don't worry.

But I must say I feel like C4 int is taught so badly. This chapter is so much understanding and so little method compared to a lot of the maths at A level. I've seen quite a few teachers just show their students the methods (like the rote way you've described, Zain) neglecting to give equal attention to when/(even why) each will work

OP not attacking you in any way, don't worry.

But I must say I feel like C4 int is taught so badly. This chapter is so much understanding and so little method compared to a lot of the maths at A level. I've seen quite a few teachers just show their students the methods (like the rote way you've described, Zain) neglecting to give equal attention to when/(even why) each will work

Because:
$-16\cos^4(x)=\displaystyle\int -64\cos^3(x)\sin(x) \ dx \not\Rightarrow-16 \dfrac{\cos^4(x)}{\sin(x)} = \displaystyle \int -64\cos^3(x) \ dx$.
The $\sin(x)$ is a function of $x$ so you can't pull it out of the integral like that.

Yes, I was having something of a mare as maybe you saw, was trying to find the error, so had split it up into two blocks of TeX, of course it was an excess bracket as usual -_-