Integrating with Trigonometric Identities: Are My Solutions Accurate?

[Natasha1]

I have been asked to find the integral (i) sinx cox dx and the integral
(ii) x sinx cosx dx using the identity sin2x = 2sinxcosx

My work...

(i) sinx cox dx

= 1/2 integral of 2 sinx cos dx

= 1/2 integral of sin 2x dx

= 1/2 cos 2x + C is this correct?

(ii) x sinx cosx dx

not sure how to do this one ?

 

[TD]

(i) sinx cox dx

= 1/2 integral of 2 sinx cos dx

= 1/2 integral of sin 2x dx

= 1/2 cos 2x + C is this correct?

Almost, you made one mistake: you forgot a minus.
(Be careful: for derivatives it's the other way arround)

(ii) x sinx cosx dx

not sure how to do this one ?

Have you seen integration by parts?

 

[HallsofIvy]

I have been asked to find the integral (i) sinx cox dx and the integral
(ii) x sinx cosx dx using the identity sin2x = 2sinxcosx

My work...

(i) sinx cox dx

= 1/2 integral of 2 sinx cos dx

= 1/2 integral of sin 2x dx

= 1/2 cos 2x + C is this correct?

Cute! Not the way I would have done but (except for the negative sign that TD mentioned) correct. What I would have done is this: let u= sin x. Then du= cos x dx so $\int sin x cos x dx= \int u du= \frac{1}{2}u^2+ C= \frac{1}{2} sin^2x+ C$ .
Can you see that that is exactly the same as your answer (again, including the missing negative)?

(ii) x sinx cosx dx

not sure how to do this one ?

As TD said, use integration by parts. Let u= x and dv= sin x cos xdx. (you already know what v is!)

 

[Natasha1]

I have been asked to find the integral (i) sinx cox dx and the integral
(ii) x sinx cosx dx using the identity sin2x = 2sinxcosx

My work...

(i) sinx cox dx

= 1/2 integral of 2 sinx cos dx

= 1/2 integral of sin 2x dx

= 1/2 cos 2x + C is this correct?

Cute! Not the way I would have done but (except for the negative sign that TD mentioned) correct. What I would have done is this: let u= sin x. Then du= cos x dx so $\int sin x cos x dx= \int u du= \frac{1}{2}u^2+ C= \frac{1}{2} sin^2x+ C$ .
Can you see that that is exactly the same as your answer (again, including the missing negative)?


As TD said, use integration by parts. Let u= x and dv= sin x cos xdx. (you already know what v is!)

Lets see if I'm right... I am doing (ii) here

u = x, v = -1/2 cos2x, du=1, dv=sinxcosx

so integral u dv = -1/2 x cos2x - [-1/2 cos 2x]
= -1/2 x cos 2x + 1/2 cos 2x + c
= 1/2 [1- x cos 2x] + C is this correct please?

 

[TD]

Lets see if I'm right... I am doing (ii) here

u = x, v = -1/2 cos2x, du=1, dv=sinxcosx

so integral u dv = -1/2 x cos2x - [-1/2 cos 2x]
= -1/2 x cos 2x + 1/2 cos 2x + c
= 1/2 [1- x cos 2x] + C is this correct please?

Didn't you forget an integral when you applied integration by parts?
It is $\int {udv = uv - } \int {vdu}$ and you left out that last integral I believe...

 

[Natasha1]

Didn't you forget an integral when you applied integration by parts?
It is $\int {udv = uv - } \int {vdu}$ and you left out that last integral I believe...

I don't think so no because from the previous example we get the integral of sin x cos x which I use at the end. So I think I am correct no ?

 

[TD]

With v = -1/2 cos(2x) and du = dx, the formula gives a new integral at the end and not just "-1/2 cos(2x)". Applying integration by parts would give:

$$\int {x\sin x\cos xdx} = - \frac{1}{2}x\cos \left( {2x} \right) - \int { - \frac{1}{2}\cos \left( {2x} \right)dx} = - \frac{1}{2}x\cos \left( {2x} \right) + \frac{1}{2}\int {\cos \left( {2x} \right)dx}$$

You forgot that last integral and just added "-1/2 cos(2x)", you see?

 

[Natasha1]

With v = -1/2 cos(2x) and du = dx, the formula gives a new integral at the end and not just "-1/2 cos(2x)". Applying integration by parts would give:

$$\int {x\sin x\cos xdx} = - \frac{1}{2}x\cos \left( {2x} \right) - \int { - \frac{1}{2}\cos \left( {2x} \right)dx} = - \frac{1}{2}x\cos \left( {2x} \right) + \frac{1}{2}\int {\cos \left( {2x} \right)dx}$$

You forgot that last integral and just added "-1/2 cos(2x)", you see?

must I have to do another integration by part on 1/2 integral of cos 2x

 

[TD]

Well, you must do one more integration but the total process is called 'integration by parts' (which you only have to do once). Have you already covered this technique? Basically, it comes down to the formula I gave two posts back.

 

[Natasha1]

With v = -1/2 cos(2x) and du = dx, the formula gives a new integral at the end and not just "-1/2 cos(2x)". Applying integration by parts would give:

$$\int {x\sin x\cos xdx} = - \frac{1}{2}x\cos \left( {2x} \right) - \int { - \frac{1}{2}\cos \left( {2x} \right)dx} = - \frac{1}{2}x\cos \left( {2x} \right) + \frac{1}{2}\int {\cos \left( {2x} \right)dx}$$

You forgot that last integral and just added "-1/2 cos(2x)", you see?

from here I do:

u = 2x
du = 2

= 1/2 integral of cos u du
= 1 / 2 sin 2x + C

so the whole thing gives - 1/2 x cos 2x + 1/2 sin 2x + C is this correct?

 

[TD]

Almost, there was already a factor 1/2 before the integral and because of your substitution, there should be another one so that would give 1/4 sin(2x) for that last part.

Then it should be correct, if you don't forget the constant of integration.
If you want to check it yourself: find its derivative and see if you get your initial function again

 

[Natasha1]

Almost, there was already a factor 1/2 before the integral and because of your substitution, there should be another one so that would give 1/4 sin(2x) for that last part.

Then it should be correct, if you don't forget the constant of integration.
If you want to check it yourself: find its derivative and see if you get your initial function again

Thanks TD you are a star!

 

[TD]

You're welcome