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#?

4 Answers
Apr 13, 2017

#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#

Apr 13, 2017

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.!

Apr 13, 2017

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.!

Apr 14, 2017

#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.!