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#?

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#?

Emilia Aguilar · Accepted Answer

#pi/2#

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#

Aurora Alvarado · Answer

Proof is given in the Explanation.

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

Charles Anctil · Answer

For the #2^(nd)# Proof, refer to the 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.!

Aurora Alvarado · Answer

#pi/2.#

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.!

Charles Anctil · Answer

undefined To prove $ u_n = \frac{2 \sin((n-1)x)}{n-1} + u_{n-2} $ for $ n \geq 2 $, we'll use mathematical induction.

Base Case: $ n = 2 $
For $ n = 2 $, we have:
$ u_2 = \int \frac{\sin(2x)}{\sin(x)} \, dx $
By trigonometric identity, $ \sin(2x) = 2\sin(x)\cos(x) $, so:
$ u_2 = \int \frac{2\sin(x)\cos(x)}{\sin(x)} \, dx = 2\int \cos(x) \, dx = 2\sin(x) + C $
This matches the form $ \frac{2\sin((2-1)x)}{2-1} + u_{2-2} $, thus the base case holds.

Inductive Step:
Assume $ u_k = \frac{2\sin((k-1)x)}{k-1} + u_{k-2} $ for some $ k \geq 2 $.
We want to prove $ u_{k+1} = \frac{2\sin(kx)}{k} + u_{k-1} $.

Consider:
$ u_{k+1} = \int \frac{\sin((k+1)x)}{\sin(x)} \, dx $
Using the angle addition formula for sine, $ \sin((k+1)x) = \sin(kx)\cos(x) + \cos(kx)\sin(x) $, we get:
$ u_{k+1} = \int \frac{\sin(kx)\cos(x) + \cos(kx)\sin(x)}{\sin(x)} \, dx $
$ = \int \cos(kx) \, dx + \int \frac{\sin(kx)}{\sin(x)} \, dx $
The first integral is $ \frac{\sin(kx)}{k} + C_1 $ and by our assumption, the second integral is $ u_k = \frac{2\sin((k-1)x)}{k-1} + u_{k-2} $.

Thus:
$ u_{k+1} = \frac{\sin(kx)}{k} + \frac{2\sin((k-1)x)}{k-1} + u_{k-2} $
Rearranging terms, we have:
$ u_{k+1} = \frac{2\sin(kx)}{k} + u_{k-1} $

This completes the inductive step.

Now, to evaluate $ \int_0^{\frac{\pi}{2}} \frac{\sin(5x)}{\sin(x)} \, dx $, we notice that $ n = 5 $ in the given formula. Using the formula derived above for $ u_5 $:
$ u_5 = \frac{2\sin(4x)}{4} + u_3 $

Now $ u_3 $ can be derived similarly using the formula. Continuing this process until we get to $ u_1 $ which is trivially $ \int \frac{\sin(x)}{\sin(x)} \, dx = x $:

$ u_5 = \frac{2\sin(4x)}{4} + u_3 $
$ = \frac{2\sin(4x)}{4} + \frac{2\sin(2x)}{2} + u_1 $
$ = \frac{2\sin(4x)}{4} + \frac{2\sin(2x)}{2} + x $

Finally, integrate this expression from $ 0 $ to $ \frac{\pi}{2} $ to get the result.

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

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#?