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Math  /  Calculus

Question∫0π3cos⁡−1(3x)dx\int_0^{\frac{\pi}{3}} \cos^{-1}(3x) dx

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of the inverse cosine of 3x3x from 00 to π3\frac{\pi}{3}. Watch out! Inverse trig functions can be tricky, so let's be careful with our substitutions and calculations!

STEP 2

1. Substitution
2. Integration by Parts
3. Definite Integral Calculation

STEP 3

Let's **begin** by making a clever substitution!
Let u=3xu = 3x.
This means du=3 dxdu = 3 \, dx, or dx=13 dudx = \frac{1}{3} \, du.

STEP 4

Now, we need to **adjust our limits of integration**.
When x=0x = 0, we have u=3⋅0=0u = 3 \cdot 0 = 0.
When x=π3x = \frac{\pi}{3}, we have u=3⋅π3=πu = 3 \cdot \frac{\pi}{3} = \pi.

STEP 5

Our integral now becomes: 13∫0πcos⁡−1(u) du \frac{1}{3} \int_0^{\pi} \cos^{-1}(u) \, du Much nicer!

STEP 6

Time for some **integration by parts**!
Remember the formula: ∫v du=uv−∫u dv\int v \, du = uv - \int u \, dv.

STEP 7

Let's choose v=cos⁡−1(u)v = \cos^{-1}(u) and du=dudu = du.
This means dv=−11−u2 dudv = -\frac{1}{\sqrt{1 - u^2}} \, du and u=uu = u.

STEP 8

Plugging into the integration by parts formula, we get: 13[ucos⁡−1(u)∣0π−∫0πu(−11−u2)du] \frac{1}{3} \left[ u \cos^{-1}(u) \Big|_0^{\pi} - \int_0^{\pi} u \left( -\frac{1}{\sqrt{1 - u^2}} \right) du \right]

STEP 9

Let's **simplify** that last integral a bit: 13[ucos⁡−1(u)∣0π+∫0πu1−u2 du] \frac{1}{3} \left[ u \cos^{-1}(u) \Big|_0^{\pi} + \int_0^{\pi} \frac{u}{\sqrt{1 - u^2}} \, du \right]

STEP 10

For the remaining integral, let's use another **substitution**!
Let w=1−u2w = 1 - u^2.
Then dw=−2u dudw = -2u \, du, so u du=−12 dwu \, du = -\frac{1}{2} \, dw.

STEP 11

When u=0u = 0, w=1w = 1.
When u=πu = \pi, w=1−π2w = 1 - \pi^2.
Our integral becomes: 13[ucos⁡−1(u)∣0π−12∫11−π21w dw] \frac{1}{3} \left[ u \cos^{-1}(u) \Big|_0^{\pi} - \frac{1}{2} \int_1^{1 - \pi^2} \frac{1}{\sqrt{w}} \, dw \right]

STEP 12

Now we can **integrate**: 13[ucos⁡−1(u)∣0π−12⋅2w∣11−π2] \frac{1}{3} \left[ u \cos^{-1}(u) \Big|_0^{\pi} - \frac{1}{2} \cdot 2\sqrt{w} \Big|_1^{1 - \pi^2} \right] 13[ucos⁡−1(u)∣0π−w∣11−π2] \frac{1}{3} \left[ u \cos^{-1}(u) \Big|_0^{\pi} - \sqrt{w} \Big|_1^{1 - \pi^2} \right]

STEP 13

Let's **evaluate**! 13[(πcos⁡−1(π)−0)−(1−π2−1)] \frac{1}{3} \left[ \left( \pi \cos^{-1}(\pi) - 0 \right) - \left( \sqrt{1 - \pi^2} - \sqrt{1} \right) \right]

STEP 14

Since cos⁡−1(π)\cos^{-1}(\pi) is not defined in the reals, we realize we made a mistake!
The original substitution u=3xu=3x changes the domain to [0,π][0, \pi], which is outside the domain of cos⁡−1(u)\cos^{-1}(u), which is [−1,1][-1, 1].
This means our approach won't work!
We need to rethink our strategy.

STEP 15

Since the domain of cos⁡−1(3x)\cos^{-1}(3x) is 3x∈[−1,1]3x \in [-1, 1], or x∈[−13,13]x \in [-\frac{1}{3}, \frac{1}{3}], and we're integrating from 00 to π3\frac{\pi}{3}, which is greater than 13\frac{1}{3}, the integral is undefined.

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