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Evaluate the integral $\begin{array}{l} I=\int_{0}^{1} \frac{x^{3}}{\sqrt{1-x^{8}}} \mathrm{~d} x \\ I= \end{array}$
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Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of a fraction involving a cube root and a square root, from 0 to 1. Watch out! Don't forget to adjust the bounds when using substitution!

STEP 2

1. Substitution
2. Simplify
3. Integrate
4. Evaluate

STEP 3

Let's make a clever substitution to simplify our integral!
We see an $x^8$ inside the square root, and an $x^3$ in the numerator.
This suggests a substitution involving $x^4$ .
Let's try $u = x^4$ .

STEP 4

Now, we need to find $\mathrm{d}u$ in terms of $\mathrm{d}x$ .
If $u = x^4$ , then $\frac{\mathrm{d}u}{\mathrm{d}x} = 4x^3$ .
So, $\mathrm{d}x = \frac{\mathrm{d}u}{4x^3}$ .
This is great because the $x^3$ will divide to one with the $x^3$ in the numerator of the original integral!

STEP 5

Don't forget to change the limits of integration!
When $x = 0$ , $u = (0)^4 = 0$ .
When $x = 1$ , $u = (1)^4 = 1$ .
So, our new limits of integration are still 0 and 1.
Lucky us!

STEP 6

Let's rewrite the integral with our substitution: $I = \int_{0}^{1} \frac{x^3}{\sqrt{1 - x^8}} \mathrm{d}x = \int_{0}^{1} \frac{x^3}{\sqrt{1 - u^2}} \cdot \frac{\mathrm{d}u}{4x^3}.$

STEP 7

The $x^3$ terms divide to one, and we can pull the constant $\frac{1}{4}$ out of the integral: $I = \frac{1}{4} \int_{0}^{1} \frac{1}{\sqrt{1 - u^2}} \mathrm{d}u.$

STEP 8

We know that the integral of $\frac{1}{\sqrt{1 - u^2}}$ is $\arcsin(u)$ .
So, $I = \frac{1}{4} \left[ \arcsin(u) \right]_0^1.$

STEP 9

Now, we just need to evaluate $\arcsin(u)$ at our limits of integration: $I = \frac{1}{4} (\arcsin(1) - \arcsin(0)).$ Since $\arcsin(1) = \frac{\pi}{2}$ and $\arcsin(0) = 0$ , we have: $I = \frac{1}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{8}.$

STEP 10

The value of the definite integral is $\frac{\pi}{8}$ .

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Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of a fraction involving a cube root and a square root, from 0 to 1. Watch out! Don't forget to adjust the bounds when using substitution!

STEP 2

1. Substitution
2. Simplify
3. Integrate
4. Evaluate

STEP 3

Let's make a clever substitution to simplify our integral!
We see an $x^8$ inside the square root, and an $x^3$ in the numerator.
This suggests a substitution involving $x^4$ .
Let's try $u = x^4$ .

STEP 4

Now, we need to find $\mathrm{d}u$ in terms of $\mathrm{d}x$ .
If $u = x^4$ , then $\frac{\mathrm{d}u}{\mathrm{d}x} = 4x^3$ .
So, $\mathrm{d}x = \frac{\mathrm{d}u}{4x^3}$ .
This is great because the $x^3$ will divide to one with the $x^3$ in the numerator of the original integral!

STEP 5

Don't forget to change the limits of integration!
When $x = 0$ , $u = (0)^4 = 0$ .
When $x = 1$ , $u = (1)^4 = 1$ .
So, our new limits of integration are still 0 and 1.
Lucky us!

STEP 6

Let's rewrite the integral with our substitution: $I = \int_{0}^{1} \frac{x^3}{\sqrt{1 - x^8}} \mathrm{d}x = \int_{0}^{1} \frac{x^3}{\sqrt{1 - u^2}} \cdot \frac{\mathrm{d}u}{4x^3}.$

STEP 7

The $x^3$ terms divide to one, and we can pull the constant $\frac{1}{4}$ out of the integral: $I = \frac{1}{4} \int_{0}^{1} \frac{1}{\sqrt{1 - u^2}} \mathrm{d}u.$

STEP 8

We know that the integral of $\frac{1}{\sqrt{1 - u^2}}$ is $\arcsin(u)$ .
So, $I = \frac{1}{4} \left[ \arcsin(u) \right]_0^1.$

STEP 9

Now, we just need to evaluate $\arcsin(u)$ at our limits of integration: $I = \frac{1}{4} (\arcsin(1) - \arcsin(0)).$ Since $\arcsin(1) = \frac{\pi}{2}$ and $\arcsin(0) = 0$ , we have: $I = \frac{1}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{8}.$

STEP 10

The value of the definite integral is $\frac{\pi}{8}$ .

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Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of a fraction involving a cube root and a square root, from 0 to 1. Watch out! Don't forget to adjust the bounds when using substitution!

STEP 2

1. Substitution2. Simplify3. Integrate4. Evaluate

STEP 3

Let's **make a clever substitution** to simplify our integral!We see an x8x^8x8 inside the square root, and an x3x^3x3 in the numerator.This suggests a substitution involving x4x^4x4.Let's try u=x4u = x^4u=x4.

STEP 4

STEP 5

Don't forget to **change the limits of integration**!When x=0x = 0x=0, u=(0)4=0u = (0)^4 = 0u=(0)4=0.When x=1x = 1x=1, u=(1)4=1u = (1)^4 = 1u=(1)4=1.So, our new limits of integration are still 0 and 1.Lucky us!

STEP 6

Let's **rewrite the integral** with our substitution: I=∫01x31−x8dx=∫01x31−u2⋅du4x3. I = \int_{0}^{1} \frac{x^3}{\sqrt{1 - x^8}} \mathrm{d}x = \int_{0}^{1} \frac{x^3}{\sqrt{1 - u^2}} \cdot \frac{\mathrm{d}u}{4x^3}. I=∫01​1−x8​x3​dx=∫01​1−u2​x3​⋅4x3du​.

STEP 7

The x3x^3x3 terms divide to one, and we can pull the constant 14\frac{1}{4}41​ out of the integral: I=14∫0111−u2du. I = \frac{1}{4} \int_{0}^{1} \frac{1}{\sqrt{1 - u^2}} \mathrm{d}u. I=41​∫01​1−u2​1​du.

STEP 8

We know that the integral of 11−u2\frac{1}{\sqrt{1 - u^2}}1−u2​1​ is arcsin⁡(u)\arcsin(u)arcsin(u).So, I=14[arcsin⁡(u)]01. I = \frac{1}{4} \left[ \arcsin(u) \right]_0^1. I=41​[arcsin(u)]01​.

STEP 9

STEP 10

The value of the definite integral is π8\frac{\pi}{8}8π​.

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1. Substitution
2. Simplify
3. Integrate
4. Evaluate

Let's make a clever substitution to simplify our integral!
We see an $x^8$ inside the square root, and an $x^3$ in the numerator.
This suggests a substitution involving $x^4$ .
Let's try $u = x^4$ .

Don't forget to change the limits of integration!
When $x = 0$ , $u = (0)^4 = 0$ .
When $x = 1$ , $u = (1)^4 = 1$ .
So, our new limits of integration are still 0 and 1.
Lucky us!

Let's rewrite the integral with our substitution: $I = \int_{0}^{1} \frac{x^3}{\sqrt{1 - x^8}} \mathrm{d}x = \int_{0}^{1} \frac{x^3}{\sqrt{1 - u^2}} \cdot \frac{\mathrm{d}u}{4x^3}.$

The $x^3$ terms divide to one, and we can pull the constant $\frac{1}{4}$ out of the integral: $I = \frac{1}{4} \int_{0}^{1} \frac{1}{\sqrt{1 - u^2}} \mathrm{d}u.$

We know that the integral of $\frac{1}{\sqrt{1 - u^2}}$ is $\arcsin(u)$ .
So, $I = \frac{1}{4} \left[ \arcsin(u) \right]_0^1.$

The value of the definite integral is $\frac{\pi}{8}$ .