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Core Maths 4: Integration/Trig/double and triple angle formulae

Hi there

Please could someone take a look at this and advise if I am on the right lines, and maybe give me a little nudge in the right direction to solve the question



Many thanks
Jackie
Reply 1
Avatar for lukas1051
lukas1051
18
No you're going down the wrong route.

Express cos^3(x) as cos(x)cos^2(x) and go from there. If you're still stuck, I'll help you some more :smile:
Reply 2
Avatar for raheem94
raheem94
13
Original post by jackie11
Hi there

Please could someone take a look at this and advise if I am on the right lines, and maybe give me a little nudge in the right direction to solve the question



Many thanks
Jackie


You are correct till cos⁡3A−3sin⁡2Acos⁡A \cos^3 A - 3 \sin^2 A \cos A

Then how did you wrote, cos⁡3A=3sin⁡2Acos⁡A ? \cos^3 A = 3 \sin^2 A \cos A \ ?

What you have is cos⁡(3A)=cos⁡3A−3sin⁡2Acos⁡A  ⟹  cos⁡3A=cos⁡(3A)+3sin⁡2Acos⁡A \cos (3A) = \cos^3 A - 3 \sin^2 A \cos A \implies \cos^3 A = \cos (3A) + 3 \sin^2 A \cos A


∫cos⁡3x dx=∫(cos⁡(3x)+3sin⁡2xcos⁡x) dx \displaystyle \int \cos^3 x \ dx = \int \left( \cos (3x) + 3 \sin^2 x \cos x \right) \ dx

Hint for integrating 3sin⁡2xcos⁡x 3 \sin^2 x \cos x

Spoiler

Reply 3
Avatar for raheem94
raheem94
13
Another approach can be to express cos⁡3x \cos^3 x as cos⁡xcos⁡2x=cos⁡x(1−sin⁡2x)=cos⁡x−cos⁡xsin⁡2x \cos x \cos^2 x = \cos x ( 1- \sin^2 x ) = \cos x - \cos x \sin^2 x
Reply 4
Avatar for Notnek
Notnek
21
Original post by Lord of the Flies
It seems you have taken the complicated route.

Try considering ∫−π/2π/2cos⁡xcos⁡2x\displaystyle\int_{-\pi/2}^{\pi/2}\cos x\cos^2 x by parts :wink:

I think your route is also complicated.

The method that Lukas gave is the best way to tackle this question.
Reply 5
Avatar for Notnek
Notnek
21
Original post by Lord of the Flies
What method?

Also, I don't see how an integration by parts is more complicated than the stream of equations in the OP:

Without difficulty, by parts twice gives:

∫cos⁡xcos⁡2x=[sin⁡xcos⁡2]−2∫sin⁡2xcos⁡x+[sin⁡3x]+2∫sin⁡2xcos⁡x\displaystyle\int\cos x\cos^2 x=[\sin x \cos^2]-2\int\sin^2 x\cos x+[\sin^3x]+2\int\sin^2x\cos x

=[sin⁡xcos⁡2]+[sin⁡3x]=[\sin x \cos^2]+[\sin^3x]

I don't mean the OP's method. I mean the method given by lukas1051:

cos⁡3x=cos⁡x(1−sin⁡2x)=cos⁡x−cos⁡xsin⁡2x\displaystyle \cos^3 x = \cos x (1-\sin^2 x) = \cos x -\cos x \sin^2 x

Both of these integrals are easy. It is far quicker and "nicer" to do this rather than use integration by parts twice.
(edited 11 years ago)
Original post by notnek
cos⁡3x=cos⁡x(1−sin⁡2x)=cos⁡x−cos⁡xsin⁡2x\displaystyle \cos^3 x = \cos x (1-\sin^2 x) = \cos x -\cos x \sin^2 x

It is far quicker and "nicer" to do this rather than use integration by parts.


Yes, just realised this... :oops:

Besides I made an idiotic mistake in my post so I'm going to leave now... :biggrin:
(edited 11 years ago)
Reply 7
Avatar for jackie11
jackie11
OP
4
Original post by raheem94
You are correct till cos⁡3A−3sin⁡2Acos⁡A \cos^3 A - 3 \sin^2 A \cos A

Then how did you wrote, cos⁡3A=3sin⁡2Acos⁡A ? \cos^3 A = 3 \sin^2 A \cos A \ ?

What you have is cos⁡(3A)=cos⁡3A−3sin⁡2Acos⁡A  ⟹  cos⁡3A=cos⁡(3A)+3sin⁡2Acos⁡A \cos (3A) = \cos^3 A - 3 \sin^2 A \cos A \implies \cos^3 A = \cos (3A) + 3 \sin^2 A \cos A



yes you are right, omg what a silly mistake to make haha :smile:
Reply 8
Avatar for jackie11
jackie11
OP
4
ok I am still not getting the right answer for this, am I on the right lines?

Original post by jackie11
ok I am still not getting the right answer for this, am I on the right lines?


∫cos⁡xsin⁡2x  dx\displaystyle\int \cos x\sin^2 x\;dx should remind you of something...

Hint:

Spoiler

Reply 10
Avatar for Notnek
Notnek
21
Original post by jackie11
ok I am still not getting the right answer for this, am I on the right lines?


You main mistake is in saying that

∫cos⁡xsin⁡2x dx=∫cosx dx×∫sin⁡2x dx \displaystyle \int \cos x \sin^2 x \ dx = \int cos x \ dx \times \int \sin^2 x \ dx

You can't split up a product in an integral like this. In general,

∫AB≠∫A×∫B\int AB \neq \int A \times \int B

Instead, you need to integrate cos⁡xsin⁡2x\cos x \sin^2 x as one expression.

Try differentiating sin⁡3x\sin^3 x and see what you get. This should give you a big clue as to what the integral of cos⁡xsin⁡2x\cos x \sin^2 x is.
(edited 11 years ago)
Reply 11
Avatar for jackie11
jackie11
OP
4
Using the chain rule, differentiating sin³ x gives me;

-3 sin² x cos x

So the integral of cos x sin² x is -1/3 sin³ x.

Am I on the right lines now? lol
Original post by jackie11
Using the chain rule, differentiating sin³ x gives me;

-3 sin² x cos x

So the integral of cos x sin² x is -1/3 sin³ x.

Am I on the right lines now? lol


Almost - you made a sign mistake. The derivative of sin is cos, not -cos.
Reply 13
Avatar for Notnek
Notnek
21
Original post by jackie11
Using the chain rule, differentiating sin³ x gives me;

-3 sin² x cos x

So the integral of cos x sin² x is -1/3 sin³ x.

Am I on the right lines now? lol

Almost, but the derivative of sin⁡3x\sin^3 x shouldn't have a negative sign since the derivative of sin⁡x\sin x is cos⁡x\cos x not −cos⁡x-\cos x.
Reply 14
Avatar for jackie11
jackie11
OP
4
ok, yes I have it!!! woohoo !!!!!!!!!!

Thank you both so much.

:smile:

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