Question

If `u_n = int (sin nx)/sinx dx, >= 2`, prove that `u_n = (2sin(n-1)x)/(n-1)+u_(n-2)` Hence evaluate: `int_0^(pi/2) (sin5x)/ sinx dx`?

Answer 1

`pi/2`

Explanation:

Calling `u_n = int sin(n x)/sinx dx` we have

`u_n-u_(n-2) = int(sin(n x)/sinx-sin((n-2) x)/sinx)dx = int 2 cos((n-1)x)dx=(2sin((n-1)x))/((n-1))`

Here we used

`sin(a+b)-sin(a-b)=2cos(a)sin(b)`

with

`a+b=nx` and
`a-b=(n-2)x`

and now

`u_5 = (2sin(4x))/4+u_3`
`u_3 = (2sin(2x))/2+u_1`

so

`u_5=[(2sin(4x))/4+ (2sin(2x))/2+x]_0^(pi/2) = pi/2`

Answer 2

Proof is given in the Explanation.

`int_0^(pi/2) sin(5x)/sinxdx=pi/2.`

Explanation:

In what follows, `n in {2,3,4,...}=NN-{1}.`

`u_n=intsin(nx)/sinxdx rArr u_(n-2)=int{sin(n-2)x}/sinxdx.`

`:. u_n-u_(n-2)=int[sin(nx)/sinx-{sin(n-2)x}/sinx]dx,`

`=int{sin(nx)-sin(nx-2x)}/sinxdx,`

`=int{2cos((nx+nx-2x)/2)*sin((nx-nx+2x)/2)}/sinx dx,`

`=2int{(cos(2nx-2x)/2)*sin(2x/2)}/sinx dx,`

`=2intcos((n-1)x)dx,`

`rArr u_n=(2sin(n-1)x)/(n-1)+u_(n-2)+C, nge2, ninNN...(star).`

Hence, the Proof.

Using `(star)` with `n=5,` we have,

`u_5=(2sin(4x))/4+u_3....(1)`

`u_3=(2sin(2x))/2+u_1....(2)`

And, finally, without `(star), u_1=intsinx/sinxdx=intdx=x...(3).`

`(1),(2),&(3) rArr u_5=1/2sin4x+sin2x+x+c.`

`:. int_0^(pi/2) sin(5x)/sinxdx=[u_5]_0^(pi/2]`

`=[1/2sin4x+sin2x+x]_0^(pi/2)`

`=[1/2sin2pi+sinpi+pi/2]-(0)`

`=pi/2.`

Enjoy Maths.!

Answer 3

For the `2^(nd)` Proof, refer to the Explanation.

Explanation:

Here is a Second Method to prove the Result for

`n>=2, n in NN.`

`u_n=intsin(nx)/sinxdx.`

Now, `sin(nx)/sinx=1/sinx{sin((nx-2x)+2x)}`

Knowing that, `sin(A+B)=sinAcosB+cosAsinB,` we get,

`sin(nx)/sinx=1/sinx{sin(nx-2x)cos2x+cos(nx-2x)sin2x}`

`=1/sinx{(sin((n-2)x))(1-2sin^2x)+(cos((n-2)x))(2sinxcosx)}`

`=1/sinx{sin((n-2)x)-2sin^2xsin((n-2)x)+2sinxcosxcos((n-2)x)}`

`=sin((n-2)x)/sinx-2sinxsin((n-2)x)+2cosxcos((n-2)x)`

`=sin((n-2)x)/sinx+2{cosxcos((n-2)x)-sinxsin((n-2)x)}`

`=sin((n-2)x)/sinx+2{cos((n-2)x+x)}`

`:. sin(nx)/sinx=sin((n-2)x)/sinx+2cos((n-1)x).`

`rArr u_n=intsin(nx)/sinxdx=int{sin((n-2)x)/sinx+2cos((n-1)x)}dx`

`=intsin((n-2)x)/sinxdx+2intcos((n-1)x)dx.`

`"Hence, "u_n=u_(n-2)+(2sin((n-1)x))/(n-1)+C; n in NN, n>=2.`

Rest of the Soln. is the same as in the First Method.

Enjoy Maths.!

Answer 4

`pi/2.`

Explanation:

Without the use of the given Recurrence Relation,

`I=int_0^(pi/2)sin(5x)/sinxdx` can be proved as shown below:

A Result :

`int_0^af(x)dx=int_0^af(a-x)dx; a>0, f :[0,a] to RR" is cont."`

We have, `I=int_0^(pi/2)sin(5x)/sinx dx............(1).`

Using the above Result,

`I=int_0^(pi/2) sin(5(pi/2-x))/{sin(pi/2-x)}dx,`

`=int_0^(pi/2) sin(5pi/2-5x)/cosxdx,`

`:. I=int_0^(pi/2) cos(5x)/cosxdx.......................(2).`

Adding `(1) and(2)`, we get,

`I+I=2I=int_0^(pi/2){sin(5x)/sinx+cos(5x)/cosx}dx,`

`=int_0^(pi/2){sin(5x)cosx+sinxcos(5x)}/(sinxcosx)dx, i.e.,`

`2I=2int_0^(pi/2){sin(5x+x)}/(2sinxcosx)dx,`

`:. I=int_0^(pi/2) sin(6x)/sin(2x)dx,`

Since, `sin6x=sin(3(2x))=3sin(2x)-4sin^3(2x)={3-4sin^2(2x)}sin(2x),`

we get, `I=int_0^(pi/2){3-4sin^2(2x)}dx,`

`=int_0^(pi/2) {3-2(1-cos4x)}dx,`

`=int_0^(pi/2)(1+2cos4x)dx,`

`=[x+2(sin(4x)/4)]_0^(pi/2),`

`=[x+1/2sin4x]_0^(pi/2),`

`={pi/2+1/2sin2pi}-{0+1/2sin0],`

`rArr I=pi/2.`

Enjoy Maths.!