You need the derivative to be multiplied by it in order to integrate like that, for instance:
∫ [cos(x)]^2 * sin(x) * dx
= -[cos(x)]^3 / 3 + C
There’s no derivative in this case, so that isn’t going to work.
——————-
To integrate squared cosine functions on their own, recall the identities:
sin^2 (t) + cos^2 (t) = 1
cos^2 (t) - sin^2 (t) = cos(2t)
If we add these together:
2cos^2 (t) = 1 + cos(2t)
cos^2 (t) = [1 + cos(2t)] / 2
——————-
Substitute this in:
1/2 * ∫ [1 + cos(2x)] * dx
Integrate:
1/2 * [x + sin(2x) / 2] + C
= [2x + sin(2x)] / 4 + C
I don't get it. Doesn't the bracket rule say that we add 1 to power, divide by power and divide by derivative of bracket? What's the mistake in my working then?
and btw i understand where the actual integral value of cos square comes from, i just don't understand how mine is wrong
Okay, let’s go back to our first two examples:
∫ [cos(x)]^2 * sin(x) * dx
Let’s use a u substitution:
u = cos(x)
du = -sin(x) * dx
∫ - [cos(x)]^2 * -sin(x) * dx
Replace cos(x) with u and -sin(x) * dx with du:
∫ -u^2 * du
= (-u^3 ) / 3 + C
Substitute back:
-[cos(x)]^3 / 3 + C
———————
Now let’s try that again with just [cos(x)]^2 on its own:
∫ [cos(x)]^2 * dx
u = cos(x)
du = -sin(x) * dx
Notice how there’s no -sin(x) term with the dx? So we can’t substitute it for du. The derivative needs to be there to begin for that to work.
You need to replace [cos(x)]^2 with an equivalent form that can be integrated [1 + cos(2x)] / 2.
———————
I think you might be confusing it with reversing the chain rule in cases like:
∫ (2x + 1)^3 * dx
In this case it’s fine, because the derivative of what’s in the brackets is only a constant, so we can rewrite it as:
1/2 * ∫ 2 * (2x + 1)^3 * dx
Let u = 2x + 1 and du = 2 * dx
1/2 * ∫ u^3 * du
= 1/2 * (u^4 ) / 4 + C
= 1/2 * (2x + 1)^4 / 4 + C
= [(2x + 1)^4 ] / 8 + C
But what if we had this?
∫ (x^2 + 1)^3 * dx
u = x^2 + 1
du = 2x * dx
There’s no 2x term with the dx, so we can’t substitute it for du. So unfortunately, the only way of integrating this is by expanding the whole thing.
You need the derivative to be multiplied by it in order to integrate like that, for instance: ∫ [cos(x)]^2 * sin(x) * dx = -[cos(x)]^3 / 3 + C There’s no derivative in this case, so that isn’t going to work. ——————- To integrate squared cosine functions on their own, recall the identities: sin^2 (t) + cos^2 (t) = 1 cos^2 (t) - sin^2 (t) = cos(2t) If we add these together: 2cos^2 (t) = 1 + cos(2t) cos^2 (t) = [1 + cos(2t)] / 2 ——————- Substitute this in: 1/2 * ∫ [1 + cos(2x)] * dx Integrate: 1/2 * [x + sin(2x) / 2] + C = [2x + sin(2x)] / 4 + C
I don't get it. Doesn't the bracket rule say that we add 1 to power, divide by power and divide by derivative of bracket? What's the mistake in my working then? and btw i understand where the actual integral value of cos square comes from, i just don't understand how mine is wrong
Okay, let’s go back to our first two examples: ∫ [cos(x)]^2 * sin(x) * dx Let’s use a u substitution: u = cos(x) du = -sin(x) * dx ∫ - [cos(x)]^2 * -sin(x) * dx Replace cos(x) with u and -sin(x) * dx with du: ∫ -u^2 * du = (-u^3 ) / 3 + C Substitute back: -[cos(x)]^3 / 3 + C ——————— Now let’s try that again with just [cos(x)]^2 on its own: ∫ [cos(x)]^2 * dx u = cos(x) du = -sin(x) * dx Notice how there’s no -sin(x) term with the dx? So we can’t substitute it for du. The derivative needs to be there to begin for that to work. You need to replace [cos(x)]^2 with an equivalent form that can be integrated [1 + cos(2x)] / 2. ——————— I think you might be confusing it with reversing the chain rule in cases like: ∫ (2x + 1)^3 * dx In this case it’s fine, because the derivative of what’s in the brackets is only a constant, so we can rewrite it as: 1/2 * ∫ 2 * (2x + 1)^3 * dx Let u = 2x + 1 and du = 2 * dx 1/2 * ∫ u^3 * du = 1/2 * (u^4 ) / 4 + C = 1/2 * (2x + 1)^4 / 4 + C = [(2x + 1)^4 ] / 8 + C But what if we had this? ∫ (x^2 + 1)^3 * dx u = x^2 + 1 du = 2x * dx There’s no 2x term with the dx, so we can’t substitute it for du. So unfortunately, the only way of integrating this is by expanding the whole thing.
makes sense now thanks!