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Math  /  Calculus

Question∫dxx324x2−324=□\int \frac{dx}{x\sqrt{324x^2 - 324}} = \square

Studdy Solution

STEP 1

What is this asking? We need to find the indefinite integral of 1x324x2−324 \frac{1}{x\sqrt{324x^2 - 324}} with respect to xx. Watch out! Don't forget the constant of integration since it's an indefinite integral!
Also, simplifying the expression first can make the integration much smoother.

STEP 2

1. Simplify the integrand.
2. Use trigonometric substitution.
3. Integrate.
4. Substitute back.

STEP 3

Alright, let's **simplify** that square root!
Notice that 324324 is a **perfect square**, and it's 18218^2!
We can factor it out: 324x2−324=324(x2−1)=182(x2−1)=18x2−1 \sqrt{324x^2 - 324} = \sqrt{324(x^2 - 1)} = \sqrt{18^2(x^2 - 1)} = 18\sqrt{x^2 - 1}

STEP 4

Now, let's rewrite the **entire integrand**: 1x324x2−324=1x⋅18x2−1=118xx2−1 \frac{1}{x\sqrt{324x^2 - 324}} = \frac{1}{x \cdot 18\sqrt{x^2 - 1}} = \frac{1}{18x\sqrt{x^2 - 1}} Much nicer, right?

STEP 5

Time for a **trig substitution**!
Since we have x2−1\sqrt{x^2 - 1}, let's set x=sec⁡(θ)x = \sec(\theta).
This is because sec⁡2(θ)−1=tan⁡2(θ) \sec^2(\theta) - 1 = \tan^2(\theta) , which will help us simplify the square root.

STEP 6

Now, we need to find dxdx.
If x=sec⁡(θ)x = \sec(\theta), then dx=sec⁡(θ)tan⁡(θ) dθdx = \sec(\theta)\tan(\theta) \, d\theta.

STEP 7

Let's **substitute** everything into our integral: ∫118xx2−1 dx=∫sec⁡(θ)tan⁡(θ)18sec⁡(θ)sec⁡2(θ)−1 dθ \int \frac{1}{18x\sqrt{x^2 - 1}} \, dx = \int \frac{\sec(\theta)\tan(\theta)}{18\sec(\theta)\sqrt{\sec^2(\theta) - 1}} \, d\theta

STEP 8

Remember that sec⁡2(θ)−1=tan⁡2(θ)\sec^2(\theta) - 1 = \tan^2(\theta), so sec⁡2(θ)−1=∣tan⁡(θ)∣\sqrt{\sec^2(\theta) - 1} = |\tan(\theta)|.
We'll assume tan⁡(θ)\tan(\theta) is positive for simplicity. ∫sec⁡(θ)tan⁡(θ)18sec⁡(θ)tan⁡(θ) dθ=∫118 dθ \int \frac{\sec(\theta)\tan(\theta)}{18\sec(\theta)\tan(\theta)} \, d\theta = \int \frac{1}{18} \, d\theta

STEP 9

This is a simple integral! ∫118 dθ=118θ+C \int \frac{1}{18} \, d\theta = \frac{1}{18}\theta + C

STEP 10

Since we let x=sec⁡(θ)x = \sec(\theta), we have θ=sec⁡−1(x)=arccos⁡(1x)\theta = \sec^{-1}(x) = \arccos(\frac{1}{x}).
Let's **substitute** this back into our result: 118θ+C=118arccos⁡(1x)+C \frac{1}{18}\theta + C = \frac{1}{18}\arccos\left(\frac{1}{x}\right) + C

STEP 11

The solution to the integral is 118arccos⁡(1x)+C\frac{1}{18}\arccos\left(\frac{1}{x}\right) + C.

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