Find the value of the definite integral

[chwala]

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,

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

Any other approach is welcome...cheers

 

[Orodruin]

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)

 

[Mark44]

On substituting the given limits, we end up with,

[1.2990−0+2.094395]−[0]=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 $\approx 0.866$.

 

[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.

 

[chwala]

Thanks...in that case i will need to leave the final answer in terms of $π$ i will look at that...

 

[chwala]

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

 

[chwala]

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,

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

Any other approach is welcome...cheers

We shall have;

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