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Integration by parts trig substitution FP2

Could anyone help me with this?

Find the integral of excosa∗cos(xsina)dxe^{xcosa} * cos(xsina) dx

Supposedly I could use a substitution and take it from there but this one seems a little tricky.
Any help appreciated!
(edited 8 years ago)

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Reply 1
Avatar for Notnek
Notnek
21
Original post by highestmountain
Could anyone help me with this?

Find the integral of excosa∗cos(xsina)dxe^{xcosa} * cos(xsina) dx

Supposedly I could use a substitution and take it from there but this one seems a little tricky.
Any help appreciated!

Is 'a' a constant? If it is then this integral can be written in a simpler way

∫ebxcos⁡cx dx\displaystyle \int e^{bx}\cos cx \ dx

And this can be integrated using parts twice with something extra that you have to do at the end. Post your working if you get stuck.
(edited 8 years ago)
Original post by notnek
Is 'a' a constant? If it is then this integral can be written in a simpler way

∫ebxcos⁡cx dx\displaystyle \int e^{bx}\cos cx \ dx

And this can be integrated using parts twice with something extra that you have to do at the end. Post your working if you get stuck.


Hi, I'm a GCSE student so I have no idea if this is right or not, so could you check the first two lines of my working please? It's here --->Integration by parts.PNG
(edited 8 years ago)
Reply 3
Avatar for Notnek
Notnek
21
Original post by Dingooose
Hi, I'm a GCSE student so I have no idea if this is right or not, so could you check the first two lines of my working please? It's here --->Integration by parts.PNG

It's nearly correct but you've lost the 'c' from the brackets of the trig function.

ddx(cos⁡cx)=−csin⁡cx\displaystyle \frac{d}{dx}\left(\cos cx\right) = -c\sin cx
Original post by notnek
It's nearly correct but you've lost the 'c' from the brackets of the trig function.

ddx(cos⁡cx)=−csin⁡cx\displaystyle \frac{d}{dx}\left(\cos cx\right) = -c\sin cx


Ah yes of course, thanks. But where would I go from there?
Reply 5
Avatar for Notnek
Notnek
21
Original post by Dingooose
Ah yes of course, thanks. But where would I go from there?

Use integration by parts again. It will feel like you're going in circles if you do it correctly. But there's a trick:

Notice that you have the same integral on the left and the right of the equation and think how you can rearrange.

Post your working if you get stuck.

Edit : Since you're not the OP, please wait for the OP to respond before posting your working. Alternatively start a new thread or pm me.
(edited 8 years ago)
Original post by Dingooose
Hi, I'm a GCSE student so I have no idea if this is right or not, so could you check the first two lines of my working please? It's here --->Integration by parts.PNG


why are you doing this level of calculus if you are a GCSE student ?
Original post by the bear
why are you doing this level of calculus if you are a GCSE student ?


I'm not OP, and I came across this thread by chance. I'm just kinda interested I guess. I certainly don't need this level of calculus as a GCSE student, but nobody is ever too young to do maths.
(edited 8 years ago)
Original post by notnek
Use integration by parts again. It will feel like you're going in circles if you do it correctly. But there's a trick:

Notice that you have the same integral on the left and the right of the equation and think how you can rearrange.

Post your working if you get stuck.

Edit : Since you're not the OP, please wait for the OP to respond before posting your working. Alternatively start a new thread or pm me.


I can't upload a pic and send it to you via PM it seems.
Reply 9
Avatar for Notnek
Notnek
21
Original post by Dingooose
I can't upload a pic and send it to you via PM it seems.

Typing it should be fine. E.g. you can write maths like this : e^x * sin(cx)
Original post by notnek
Typing it should be fine. E.g. you can write maths like this : e^x * sin(cx)


Thanks, I sent you a PM.
Original post by notnek
Is 'a' a constant? If it is then this integral can be written in a simpler way

∫ebxcos⁡cx dx\displaystyle \int e^{bx}\cos cx \ dx

And this can be integrated using parts twice with something extra that you have to do at the end. Post your working if you get stuck.


∫ebxcos⁡cx dx\displaystyle \int e^{bx}\cos cx \ dx = 1b\frac{1}{b} ebxe^{bx} ?
Original post by highestmountain
∫ebxcos⁡cx dx\displaystyle \int e^{bx}\cos cx \ dx = 1b\frac{1}{b} ebxe^{bx} ?


Unparseable latex formula:

\displaystyle \int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$$



Let g′(x)=ebx g'(x) = e^{bx}
And let f(x)=cos(cx) f(x) = cos(cx)

See what happens when you plug them in.
(edited 8 years ago)
Original post by Dingooose
Unparseable latex formula:

\displaystyle \int f(x)g'(x)=f(x)g(x)-\int f'(x)g(x)$$



Let g′(x)=ebx g'(x) = e^{bx}
And let f(x)=cos(cx) f(x) = cos(cx)

See what happens when you plug them in.


I integrated by parts twice over but can only seem to get that far after recognising that the integral appears twice - I rearranged the expression to get a fraction of the integral on its own and then divided both sides by that fraction and that's what I've got? :s-smilie:
Original post by highestmountain
I integrated by parts twice over but can only seem to get that far after recognising that the integral appears twice - I rearranged the expression to get a fraction of the integral on its own and then divided both sides by that fraction and that's what I've got? :s-smilie:


Notnek said he's going for a while in a PM, so it looks like we've only got each other. Let's try to solve this by ourselves.
Original post by highestmountain
I integrated by parts twice over but can only seem to get that far after recognising that the integral appears twice - I rearranged the expression to get a fraction of the integral on its own and then divided both sides by that fraction and that's what I've got? :s-smilie:


First integration by parts:
∫ebxcos(cx)dx=1bebxcos(cx)+cb∫sin(cx)ebxdx \displaystyle \int e^{bx}cos(cx)dx = \frac {1} {b} e^{bx}cos(cx) + \frac {c} {b} \int sin(cx)e^{bx}dx

Second integration by parts:
∫sin(cx)ebxdx=1bebxsin(cx)−cb∫cos(cx)ebxdx \displaystyle \int sin(cx)e^{bx}dx = \frac {1} {b} e^{bx}sin(cx) - \frac {c} {b} \int cos(cx)e^{bx}dx

Are we on the same page so far?
Reply 16
Avatar for Notnek
Notnek
21
Original post by Dingooose
First integration by parts:
∫ebxcos(cx)dx=1bebxcos(cx)+cb∫sin(cx)ebxdx \displaystyle \int e^{bx}cos(cx)dx = \frac {1} {b} e^{bx}cos(cx) + \frac {c} {b} \int sin(cx)e^{bx}dx

Second integration by parts:
∫sin(cx)ebxdx=1bebxsin(cx)−cb∫cos(cx)ebxdx \displaystyle \int sin(cx)e^{bx}dx = \frac {1} {b} e^{bx}sin(cx) - \frac {c} {b} \int cos(cx)e^{bx}dx

Are we on the same page so far?

Substitute for the integral using your second equation into the first equation.

You should end up with an equation that only contains the integral ∫ebxcos(cx) dx\int e^{bx} cos(cx) \ dx, which should appear twice.

By the way this integral is beyond A Level Maths i.e. wouldn't be in an exam.
(edited 8 years ago)
Original post by notnek
Substitute for the integral using your second equation into the first equation.

You should end up with an equation that only contains the integral ∫ebxcos(cx) dx\int e^{bx} cos(cx) \ dx, which should appear twice.

By the way this integral is beyond A Level Maths i.e. wouldn't be in an exam.


Do you mean this?

∫ebxcos(cx)dx=1bebxcos(cx)+cb(1bebxsin(cx)−cb∫cos(cx)ebxdx) \displaystyle \int e^{bx}cos(cx)dx = \frac {1} {b} e^{bx}cos(cx) + \frac {c} {b} \left(\frac {1} {b} e^{bx}sin(cx) - \frac {c} {b} \int cos(cx)e^{bx}dx\right)

Spoiler

(edited 8 years ago)
Original post by notnek
Substitute for the integral using your second equation into the first equation.

You should end up with an equation that only contains the integral ∫ebxcos(cx) dx\int e^{bx} cos(cx) \ dx, which should appear twice.

By the way this integral is beyond A Level Maths i.e. wouldn't be in an exam.


I THINK I FINALLY DID IT!!! See my previous reply with the spoiler tag.
(edited 8 years ago)

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