# Find the value of the definite integral

Gold Member
Homework Statement:
see attached
Relevant Equations:
integration
Find question here, My approach, using cosine sum and product concept, we shall have;

##\cos (A+B)-\cos (A-B)=-2\sin A\sin B##

##⇒\cos D-\cos C=-2\sin\dfrac{C+D}{2} \sin\dfrac {C-D}{-2}##

##⇒-3[\cos(A+B)-\cos(A-B)]=6\sin A sinB##

We are given ##A=4θ## and ##B=2θ##, therefore,

##⇒-3[\cos 6θ-\cos 2θ]=6\sin 4θ\sin2θ##

##⇒3\cos 2θ-3\cos 6θ=6\sin 4θ\sin2θ##

I made use of chain rule in my working... i.e letting ##u=2θ## and ##u=6θ## ...

##\int_0^\frac{π}{3} [3\cos 2θ-3\cos 6θ+2] dθ=\dfrac{3}{2} \sin 2θ-\dfrac{1}{2} \sin 6θ +2θ## from ##[θ=0]## to ##[θ=\frac{π}{3}]##

On substituting the given limits, we end up with,

##[1.2990-0+2.094395]-=3.393## to 3 decimal places...

Any other approach is welcome...cheers

Last edited:
• Delta2

Staff Emeritus
Homework Helper
Gold Member
Giving something to three decimal places is not giving the exact value. (Unless the decimal expansion terminates there - which it doesn’t in this case)

• chwala and Mark44
Mentor
On substituting the given limits, we end up with,

[1.2990−0+2.094395]−=3.393 to 3 decimal places...
As @Orodruin pointed out, your answer doesn't comply with the problem's request to "find the exact value".

For example, if ##\theta = \frac \pi 3##, then the value of ##\sin(2\theta)## is ##\sin(\frac{2\pi}3)##. As an exact value, this is ##\frac{\sqrt 3}2##; as an approximate value this is ##\approx 0.866##.

• chwala
Fred Wright
Another approach would be make the substitution ##u=2\theta## and observe,
$$sin(2u)sin(u)=2\cos(u)\sin^2(u)$$
A further substitution should immediately come to mind.

• Delta2 and SammyS
Gold Member
Thanks...in that case i will need to leave the final answer in terms of ##π## i will look at that...

Gold Member
Aaaaargh this was easy...I will post the exact solution later in the day...when I get hold of my laptop...

Gold Member
Homework Statement:: see attached
Relevant Equations:: integration

Find question here,
View attachment 301958

My approach, using cosine sum and product concept, we shall have;

##\cos (A+B)-\cos (A-B)=-2\sin A\sin B##

##⇒\cos D-\cos C=-2\sin\dfrac{C+D}{2} \sin\dfrac {C-D}{-2}##

##⇒-3[\cos(A+B)-\cos(A-B)]=6\sin A sinB##

We are given ##A=4θ## and ##B=2θ##, therefore,

##⇒-3[\cos 6θ-\cos 2θ]=6\sin 4θ\sin2θ##

##⇒3\cos 2θ-3\cos 6θ=6\sin 4θ\sin2θ##

I made use of chain rule in my working... i.e letting ##u=2θ## and ##u=6θ## ...

##\int_0^\frac{π}{3} [3\cos 2θ-3\cos 6θ+2] dθ=\dfrac{3}{2} \sin 2θ-\dfrac{1}{2} \sin 6θ +2θ## from ##[θ=0]## to ##[θ=\frac{π}{3}]##

On substituting the given limits, we end up with,

##[1.2990-0+2.094395]-=3.393## to 3 decimal places...

Any other approach is welcome...cheers
We shall have;

##...\dfrac{3}{2} \sin \dfrac{2π}{3}-\dfrac{1}{2} \sin 2π+\dfrac{2π}{3}##
##=\dfrac{3}{2}⋅\dfrac{\sqrt 3}{2}+\dfrac{2π}{3}##
##=\dfrac{3\sqrt 3}{4}+\dfrac{2π}{3}##

• Delta2