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# Integration watch

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1. Hi,

How do you do ∫ 2^x dx? And what is the answer?

Thanks!
2. Let y = a^x
Taking logs of both sides,
ln y = ln a^x
ln y = x ln a
Differentiating (using chain rule on left hand side),
1/y * dy/dx = ln a (because x ln a is just a constant multiple of x, like 3x)
But y = a^x.
1 / a^x * dy/dx = ln a
dy/dx = a^x ln a.

So if you differentiate 2^x, you'll get 2^x ln 2. Integration is the opposite of differentiation, and so integrating 2^x you'll get (2^x) / ln2 (+ C).
3. I know that if you differentiate 2^x you get 2^xln2, but I can't see how integrating 2^x gives (2^x) / ln2 (+ C). Is it that to differentiate it you times by ln2 so to integrate it you divide by ln2? Can you do that?
4. (Original post by sweet_gurl)
I know that if you differentiate 2^x you get 2^xln2, but I can't see how integrating 2^x gives (2^x) / ln2 (+ C). Is it that to differentiate it you times by ln2 so to integrate it you divide by ln2? Can you do that?
Yeah, more or less. Say you had a function y of x such that dy/dx = 2^x. It's obvious that y = k.2^x (some constant multiple), because to differentiate you just multiply by ln 2.

d/dx k.2^x = k d/dx 2^x = k . ln2 . 2^x
but we said that dy/dx = 2^x.
k . ln2 . 2^x = 2^x
k = 1/ln2
y = 2^x / ln2.

It's just using the fact that integration and differentiation are opposites. d/dx e^2x = 2e^2x, therefore ∫ e^2x dx = 1/2 e^2x.
5. Ahh, I see. Thank you very much for your help!
6. I've got another question:

How would you integrate (sinx)^3 dx ?
7. You could use (If you do Further Maths)

cos^3x = [1/2(z+1/z)]^3
=>cos^3x = 1/8[z^3 + 3z + 3/z + 1/z^3]
=>cos^3x = 1/8[(z^3 + 1/z^3) +3(z+1/z)]
=>cos^3x = 1/8[2cos3x + 6cosx]

which you can integrate easily.
8. split sin^3x into sinx(sin^2x)

rearrange sin^2x into 1-cos^2x

Mulitply out and integrate

In the spoiler is a hint if you get stuck.

Spoiler:
Show
remember
9. Thanks! My answer includes a (cosx)^3 term. Does this matter, or do you have to simplify it further?
10. (Original post by sweet_gurl)
Thanks! My answer includes a (cosx)^3 term. Does this matter, or do you have to simplify it further?
I think cos3x is fairly simple. Do you mean do you have to convert it to sinx? Nah.
11. If you want another way to integrate use so

Then you just use . Saves having to do the whole differentiation thing.
12. Yep thats how i would do it.

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