Question

The integral of 5xsqrt(1-x^4) = ?

Answer 1

\[\int\limits_{}^{}5x \sqrt{1-x^4}dx\]

Answer 2

Please help. Ive tried Trig Sub but im doing something incorrect

Answer 3

Hmmm, not completely sure myself.

Answer 4

What if you let \[
x^2=\sin(\theta)
\]

Answer 5

I'll show you what I attempted.

Answer 6

\[\int\limits_{}^{}5x \sqrt{1-x^4}dx\]
\[5\int\limits_{}^{}x \sqrt{1-x^4}dx\]
\[5\int\limits_{}^{}x \sqrt{1-(x^2)^2}dx\]
\[5\int\limits_{}^{}x \sqrt{1-\sin^2(\theta})^2dx\]

Answer 7

Nope.

Answer 8

Do you know how to do a substitution such as \(x^2=\sin(\theta)\)?

Answer 9

Not sure.

Answer 10

if x squared = sin(theta) the radical would be 1-sin^2(theta), right?

Answer 11

Differentiate both sides with respect to \(\theta\) to get: \[
2x\frac{dx}{d\theta}=\cos(\theta)
\]

Answer 12

so \[1/2(5)\int\limits_{}^{}(\sqrt{\sin(\theta)}\sqrt{1-\sin^2(\theta)} )d( \theta)\]

Answer 13

We can simplify this: \[
\frac{dx}{d\theta} = \frac{\cos(\theta)}{2x}\implies dx=\frac{\cos(\theta)}{2x}d\theta \implies 2x\;dx=\cos(\theta) \;d\theta
\]

Answer 14

So \[
\frac{5}{2}\int \sqrt{1-(x^2)^2}\cdot 2x\;dx =\frac{5}{2}\int\sqrt{1-\sin^2(\theta)}\;d\theta
\]

Answer 15

So \[5/2\int\limits_{}^{}(\sqrt{\cos^2(\theta)})d(\theta)\]

Answer 16

Yes

Answer 17

\[5/2\int\limits_{}^{}\cos(\theta)d(\theta)\]

Answer 18

Should be \(|\cos(\theta)|\)

Answer 19

\[5/2(\sin(\theta))+C\]

Answer 20

so \[5/2x^2+C\]

Answer 21

?

Answer 22

Ummm, does it make sense when you differentiate it?

Answer 23

It doesnt make sense and Im confused.

Answer 24

Hmm, if we did \(x=\sqrt{\sin(\theta)}\) maybe...

Answer 25

I'm not sure, let me think.

Answer 26

I appreciate it.

Answer 27

Let's start with \(u=x^2\) so \(du=2xdx\)\[
\frac{5}{2}\int\sqrt{1-(x^2)^2}2xdx = \frac{5}{2}\int\sqrt{1-u^2}du
\]

Answer 28

Then we do \(u = \sin\theta\) so \(du=\cos\theta d\theta\)\[
\frac{5}{2}\int\sqrt{1-u^2}du=\frac 52\int \sqrt{1-\sin^2\theta}\cos\theta d\theta
\]

Answer 29

That gives us \[
\frac52\int \cos^3\theta d\theta
\]

Answer 30

Right, Im following.

Answer 31

I should say \[
\frac 52\int \cos^2\theta d\theta
\]

Answer 32

then \[
\frac54\int 1+\cos(2\theta)d\theta
\]

Answer 33

Ohh, I see.

Answer 34

I found the error with my previous substitution. I put: \[
\frac{5}{2}\int \sqrt{1-(x^2)^2}\cdot 2x\;dx =\frac{5}{2}\int\sqrt{1-\sin^2(\theta)}\;d\theta
\]It should have been\[
\frac{5}{2}\int \sqrt{1-(x^2)^2}\cdot 2x\;dx =\frac{5}{2}\int\sqrt{1-\sin^2(\theta)}\;\color{red}{\cos(\theta)}d\theta
\]

Answer 35

How did we not see that? You're right. So let u = sin theta and du = cos theta. Then 5/2 integral of rad (1-u^2)du ?

Answer 36

Well \(x^2=\sin(\theta)\) or \(x^2=u=\sin(\theta)\) both lead to the same result.

Answer 37

Correct.

Answer 38

Just continue where I left off.

Answer 39

so 5/2 integral of (cos theta)^2

Answer 40

Use half angle.

Answer 41

(1+cos(2 theta))/2 ?

Answer 42

yes

Answer 43

Thank you very much @wio