# Integrate #intx^3/sqrt(x^2+4)# using trig substitution?

See the explanatiom below

You have to change as follows

It's easier without trigonometric substitution

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To integrate (\int \frac{x^3}{\sqrt{x^2 + 4}}) using trigonometric substitution, perform the following steps:

- Let (x = 2\tan(\theta)).
- Find (\frac{dx}{d\theta}) and express (x^2) in terms of (\theta).
- Substitute (x) and (dx) into the integral.
- Express (x^3) in terms of (\theta).
- Replace (x^3) and (\sqrt{x^2 + 4}) with expressions involving (\theta).
- Simplify the resulting integral in terms of (\theta).
- Integrate with respect to (\theta).
- Substitute back the value of (x) in terms of (\theta).
- Simplify the final expression.

Here are the steps in detail:

- Let (x = 2\tan(\theta)).
- (dx = 2\sec^2(\theta)d\theta).
- (x^2 = (2\tan(\theta))^2 = 4\tan^2(\theta) = 4(\sec^2(\theta) - 1) = 4\sec^2(\theta) - 4).
- (x^3 = (2\tan(\theta))^3 = 8\tan^3(\theta)).
- (\sqrt{x^2 + 4} = \sqrt{4\sec^2(\theta) - 4} = 2\sqrt{\sec^2(\theta) - 1} = 2\tan(\theta)).
- Substitute (x^3) and (\sqrt{x^2 + 4}) into the integral: (\int \frac{8\tan^3(\theta)}{2\tan(\theta)} \cdot 2\sec^2(\theta)d\theta).
- Simplify: (\int 4\tan^2(\theta)\sec^2(\theta)d\theta).
- Integrate with respect to (\theta): (= \int 4(\sec^4(\theta) - 1)d\theta).
- (= 4\int \sec^4(\theta)d\theta - 4\int d\theta).
- The integral of (\sec^4(\theta)) can be evaluated using integration by parts.
- Let (u = \sec^2(\theta)) and (dv = \sec^2(\theta)d\theta).
- Find (du) and (v), then apply the integration by parts formula.
- After integration by parts, simplify the expression.
- Integrate (4\int d\theta).
- Substitute back the value of (x) in terms of (\theta) and simplify the final expression.

This completes the integration of (\int \frac{x^3}{\sqrt{x^2 + 4}}) using trigonometric substitution.

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