How to Solve the Integral of \( \sin^3(x) \cos^2(x) \)?

[shreddinglicks]

Homework Statement

25∫5(sin(x))^3*cos(x)^2 dx

Homework Equations

25∫5(sin(x))^3*cos(x)^2 dx

The Attempt at a Solution

5(sin(x))^2 * sin(x) * (cos(x))^2
5(1-(cos(x))^2) * sin(x) * (cos(x))^2

25∫5((cos(x)^2)-(cos(x)^4))sin(x) dx

Now I'm stuck.

 

[Mentallic]

Just make a certain u-substitution, and everything will fall into place.

 

[shreddinglicks]

Just make a certain u-substitution, and everything will fall into place.

ok so I do

u = cos(x)
du = -sin(x)

the part I am confused about is do I multiply the 5 with 25 to get 125?

giving me

-125( (u^3/3) - (u^5/5)) + C

The integral calculator I used says its completely wrong.

 

[shreddinglicks]

I should have

([25((cosx)^3)(3cos(2x)-7)] / [6] )+ C

I don't have any clue how to get that answer.

 

[shreddinglicks]

This integral is actually part of a bigger problem regarding trig. u-sub.

∫(x^3)sqrt(25-x^2) dx

I'm just stuck at the step that you see in the earlier post.

 

[Mentallic]

Yes, you can move that 5 in the integral outside to get 125.

So what you have after integrating with respect to u is

[tex]125\left(\frac{u^3}{3}-\frac{u^5}{5}\right)+C[/tex]

which is correct. After substituting back to get it in terms of x,

[tex]125\left(\frac{\cos^3{x}}{3}-\frac{\cos^5{x}}{5}\right)+C[/tex]

Which is a perfectly valid answer, but if you want it to look like what your integral calculator gave you, notice that its answer is of the form

[itex]k\cdot\cos^3{x}(...)+c[/itex]

which means if we want it to look like that, we need to factor out [itex]\cos^3{x}[/itex] and then work on the other factor (...) that's left over.

 

[Mentallic]

In case you're still confused with regards to the 5 multiplied by 25, if you ever have an integral of the form

[tex]a_1\int{a_2\cdot f(x)dx}[/tex]

(where a₁ and a₂ are some constants)

then you can always move the a₂ out of the integral to get

[tex]a_1a_2 \int {f(x)dx}[/tex]

which, after integrating we get

[tex]a_1a_2\left[F(x)+c\right][/tex]

expanded,

[tex]a_1a_2F(x)+a_1a_2c[/tex]

But now let's say you didn't take a₂ out the front and integrated directly, then you'd get

[tex]a_1\left[a_2F(x)+c\right][/tex]

expanded,

[tex]a_1a_2F(x)+a_1c[/tex]

which doesn't seem to be the same as before. It isn't, but that's the point of the constant of integration. If we had to solve a certain problem, then the c in our first integration would be different to the c in our second integration such that we would end up getting the same answer. The moral of the story is, if your integral calculator says something that you're not getting, keep in mind that its constant of integration could be different to yours.

I've been in a couple of situations myself where I would get an answer like

[tex]\tan{x}+\tan^{-1}{(1/2)}+c[/tex]

while wolfram alpha would give the answer

[tex]\tan{x}+c[/tex]

and I'd go back and forth indefinitely, furiously trying to figure out why I have this extra value in my answer.

 

[shreddinglicks]

Yes, you can move that 5 in the integral outside to get 125.

So what you have after integrating with respect to u is

[tex]125\left(\frac{u^3}{3}-\frac{u^5}{5}\right)+C[/tex]

which is correct. After substituting back to get it in terms of x,

[tex]125\left(\frac{\cos^3{x}}{3}-\frac{\cos^5{x}}{5}\right)+C[/tex]

Which is a perfectly valid answer, but if you want it to look like what your integral calculator gave you, notice that its answer is of the form

[itex]k\cdot\cos^3{x}(...)+c[/itex]

which means if we want it to look like that, we need to factor out [itex]\cos^3{x}[/itex] and then work on the other factor (...) that's left over.

I see, thanks for clearing that up.

I must have some error with my algebra then. My final answer for my trig. u-sub. problem is wrong.

for:

∫(x^3)sqrt(25-x^2) dx

I know have:

-125*((5*((1/5)*sqrt(25-x^2))^3-3*((1/5)*sqrt(25-x^2))^5)*(1/15))

when I simplify this my answer is wrong.

 

[shreddinglicks]

In case you're still confused with regards to the 5 multiplied by 25, if you ever have an integral of the form

[tex]a_1\int{a_2\cdot f(x)dx}[/tex]

(where a₁ and a₂ are some constants)

then you can always move the a₂ out of the integral to get

[tex]a_1a_2 \int {f(x)dx}[/tex]

which, after integrating we get

[tex]a_1a_2\left[F(x)+c\right][/tex]

expanded,

[tex]a_1a_2F(x)+a_1a_2c[/tex]

But now let's say you didn't take a₂ out the front and integrated directly, then you'd get

[tex]a_1\left[a_2F(x)+c\right][/tex]

expanded,

[tex]a_1a_2F(x)+a_1c[/tex]

which doesn't seem to be the same as before. It isn't, but that's the point of the constant of integration. If we had to solve a certain problem, then the c in our first integration would be different to the c in our second integration such that we would end up getting the same answer. The moral of the story is, if your integral calculator says something that you're not getting, keep in mind that its constant of integration could be different to yours.

I've been in a couple of situations myself where I would get an answer like

[tex]\tan{x}+\tan^{-1}{(1/2)}+c[/tex]

while wolfram alpha would give the answer

[tex]\tan{x}+c[/tex]

and I'd go back and forth indefinitely, furiously trying to figure out why I have this extra value in my answer.

Yes, I agree it was beating me up for awhile.

 

[shreddinglicks]

I'm not sure about your location. I'm on the east coast of North America and it's 1am. I have to be up at 8am to take care of things. If you post I may not reply till sometime later in the day.

Thanks for your help so far. I will work on this problem again as soon as I get the chance.

 

[Mentallic]

I see, thanks for clearing that up.

I must have some error with my algebra then. My final answer for my trig. u-sub. problem is wrong.

for:

∫(x^3)sqrt(25-x^2) dx

I know have:

-125*((5*((1/5)*sqrt(25-x^2))^3-3*((1/5)*sqrt(25-x^2))^5)*(1/15))

when I simplify this my answer is wrong.

How does your original problem relate to this integral that you have to solve? For such an integral, you should be looking to make a u-substitution such that the square root will be gone. Using

[tex]x=5\sin{u}[/tex]
or
[tex]x=5\cos{u}[/tex]
will achieve this

I'm not sure about your location. I'm on the east coast of North America and it's 1am. I have to be up at 8am to take care of things. If you post I may not reply till sometime later in the day.

Thanks for your help so far. I will work on this problem again as soon as I get the chance.

No worries. Take your time!

 

[shreddinglicks]

I figured I'll send this real quick before I turn off the PC.

This is the work I have done.

 

[shreddinglicks]

Sorry, I just realized it's upside down. Didn't mean to do that.

If you have issues with it let me know. I'll resend 1st chance I get.

 

[Mentallic]

The mistake you made was turning

[tex]125\sin^3{\theta}\times 25\cos^2{\theta}[/tex]

into

[tex]25\left(5\sin^3{\theta}\times\cos^2{\theta}\right)[/tex]

You're factorizing as though we're adding two values, but we're not, we're multiplying.[tex]125\sin^3{\theta}\times 25\cos^2{\theta}[/tex]

[tex]=125\times 25 \times\sin^3{\theta}\cos^2{\theta}[/tex]

[tex]=5^5\sin^3{\theta}\cos^2{\theta}[/tex]

If you have something like

[tex]a\cdot x\cdot a\cdot y[/tex]

then this is equivalent to

[tex]a^2xy[/tex]

then if you factor out 'a', you'll only be taking out one of the factors of a, hence [itex]axay=a(xay)[/itex] and not [itex]a(xy)[/itex]

 

[shreddinglicks]

I did this in hast, Can you please check it over?

 

[shreddinglicks]

Wait, I think I should've done all the work in the parenthesis first. Sorry things are hectic. I'm trying to do my homework and meanwhile my mother could care less and has me scrambling to go visit some relatives for the holiday.

I'll redo and re post as soon as I can.

 

[Mentallic]

It's pretty hard to follow what you've done to be honest. At the point where you substitute f(x) back in for u, you've divided the entire expression by 15 while the numerators didn't change? C also seems to come and go as it pleases. When you're in less of a rush, be sure that each line follows logically from the last.

 

[shreddinglicks]

I actually got the correct answer while sitting on the train yesterday. I'll be heading home tomorrow afternoon. When I get home I'll scan my work so you can see it.

Once again, thanks for your help. I really appreciate it.

 

[Mentallic]

You've integrated correctly so your answer is correct. If you were then able to express it in the same way as the answer then it's likely you're ok, unless you've made at least two algebraic errors that both just happened to cancel each other out

 

[shreddinglicks]

This is the completed work.

 

[Mentallic]

Looks good! And I applaud your efforts to make it neat. You have no idea how much we've had to try and decipher unreadable uploads.