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Question: If we have an integral as \[{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + {\pi ^x}} \ri...

If we have an integral as In=∫−ππsin⁡nx(1+πx)sin⁡xdx,n=0,1,2,3.....{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx,n = 0,1,2,3.....} then
(a)In=In+2\left( a \right){I_n} = {I_{n + 2}}
(b)∑m=110I2m+1=10π\left( b \right)\sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = 10\pi
(c)∑m=110I2m=0\left( c \right)\sum\limits_{m = 1}^{10} {{I_{2m}}} = 0
(d)In=In+1\left( d \right){I_n} = {I_{n + 1}}

Explanation

Solution

In this particular question use the property of definite integral i.e. ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} so apply this and after add both of the integration so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given integral:
In=∫−ππsin⁡nx(1+πx)sin⁡xdx{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx} ....................... (1)
Now according to the definite integral property ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} we have,
⇒In=∫−ππsin⁡n(−π+π−x)(1+π(−π+π−x))sin⁡(−π+π−x)dx\Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin n\left( { - \pi + \pi - x} \right)}}{{\left( {1 + {\pi ^{\left( { - \pi + \pi - x} \right)}}} \right)\sin \left( { - \pi + \pi - x} \right)}}dx}
Now simplify this using the property that (sin (-x) = -sin x) so we have,
⇒In=∫−ππsin⁡n(−x)(1+π(−x))sin⁡(−x)dx\Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin n\left( { - x} \right)}}{{\left( {1 + {\pi ^{\left( { - x} \right)}}} \right)\sin \left( { - x} \right)}}dx}
⇒In=∫−ππ−sin⁡nx−(1+π(−x))sin⁡xdx\Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{ - \sin nx}}{{ - \left( {1 + {\pi ^{\left( { - x} \right)}}} \right)\sin x}}dx}
⇒In=∫−ππsin⁡nx(1+1πx)sin⁡xdx\Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + \dfrac{1}{{{\pi ^x}}}} \right)\sin x}}dx}
⇒In=∫−πππxsin⁡nx(πx+1)sin⁡xdx\Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{{\pi ^x}\sin nx}}{{\left( {{\pi ^x} + 1} \right)\sin x}}dx}....................... (2)
Now add equation (1) and (2) we have,
⇒2In=∫−ππsin⁡nx(1+πx)sin⁡xdx+∫−πππxsin⁡nx(πx+1)sin⁡xdx\Rightarrow 2{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx} + \int_{ - \pi }^\pi {\dfrac{{{\pi ^x}\sin nx}}{{\left( {{\pi ^x} + 1} \right)\sin x}}dx}
⇒2In=∫−ππ(1+πx)sin⁡nx(1+πx)sin⁡xdx\Rightarrow 2{I_n} = \int_{ - \pi }^\pi {\dfrac{{\left( {1 + {\pi ^x}} \right)\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx}
⇒2In=∫−ππsin⁡nxsin⁡xdx\Rightarrow 2{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\sin x}}dx}
⇒In=12∫−ππsin⁡nxsin⁡xdx\Rightarrow {I_n} = \dfrac{1}{2}\int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\sin x}}dx}
Now as we know that ∫−aaf(x)dx=2∫0af(x)dx\int_{ - a}^a {f\left( x \right)dx} = 2\int_0^a {f\left( x \right)dx} , if and only if f (x) is an even function.
So, as we know that sin (x) is an odd function so the ratio of odd function to the odd function is an even function, so sin⁡nxsin⁡x=\dfrac{{\sin nx}}{{\sin x}} = even function.
⇒In=12∫−ππsin⁡nxsin⁡xdx=22∫0πsin⁡nxsin⁡xdx\Rightarrow {I_n} = \dfrac{1}{2}\int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\sin x}}dx} = \dfrac{2}{2}\int_0^\pi {\dfrac{{\sin nx}}{{\sin x}}dx}
⇒In=∫0πsin⁡nxsin⁡xdx\Rightarrow {I_n} = \int_0^\pi {\dfrac{{\sin nx}}{{\sin x}}dx}................ (3)
Now substitute in place of n, (n + 2) so we have,
⇒In+2=∫0πsin⁡(n+2)xsin⁡xdx\Rightarrow {I_{n + 2}} = \int_0^\pi {\dfrac{{\sin \left( {n + 2} \right)x}}{{\sin x}}dx}................ (4)
Now subtract equation (3) from equation (4) we have,
⇒In+2−In=∫0πsin⁡(n+2)x−sin⁡nxsin⁡xdx\Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {\dfrac{{\sin \left( {n + 2} \right)x - \sin nx}}{{\sin x}}dx}
Now as we know that sin C – sin D = 2sin⁡(C−D2)cos⁡(C+D2)2\sin \left( {\dfrac{{C - D}}{2}} \right)\cos \left( {\dfrac{{C + D}}{2}} \right) so use this property we have,
⇒In+2−In=∫0π2sin⁡(nx+2x−nx2)cos⁡(nx+2x+nx2)sin⁡xdx\Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {\dfrac{{2\sin \left( {\dfrac{{nx + 2x - nx}}{2}} \right)\cos \left( {\dfrac{{nx + 2x + nx}}{2}} \right)}}{{\sin x}}dx}
⇒In+2−In=∫0π2sin⁡xcos⁡(n+1)xsin⁡xdx\Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {\dfrac{{2\sin x\cos \left( {n + 1} \right)x}}{{\sin x}}dx}
⇒In+2−In=∫0π2cos⁡(n+1)xdx\Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {2\cos \left( {n + 1} \right)xdx}
Now integrate it we have,
⇒In+2−In=[2sin⁡(n+1)xn+1]0π\Rightarrow {I_{n + 2}} - {I_n} = \left[ {\dfrac{{2\sin \left( {n + 1} \right)x}}{{n + 1}}} \right]_0^\pi
Now apply integrating limits we have,
⇒In+2−In=[2sin⁡(n+1)πn+1−2sin⁡0n+1]\Rightarrow {I_{n + 2}} - {I_n} = \left[ {\dfrac{{2\sin \left( {n + 1} \right)\pi }}{{n + 1}} - \dfrac{{2\sin 0}}{{n + 1}}} \right]
Now as it is given that n = 0, 1, 2, 3.... so n is an integer so, sin⁡(n+1)π=0\sin \left( {n + 1} \right)\pi = 0 and sin 0 = 0 so we h ave,
⇒In+2−In=[0−0]=0\Rightarrow {I_{n + 2}} - {I_n} = \left[ {0 - 0} \right] = 0
⇒In=In+2\Rightarrow {I_n} = {I_{n + 2}}............... (5)
So option (a) is correct.
Now in equation (3) substitute n = 1 we have,
⇒I1=∫0πsin⁡xsin⁡xdx=∫0π1.dx=[x]0π=π\Rightarrow {I_1} = \int_0^\pi {\dfrac{{\sin x}}{{\sin x}}dx} = \int_0^\pi {1.dx} = \left[ x \right]_0^\pi = \pi..................... (6)
Now in equation (3) substitute n = 2 we have,
⇒I2=∫0πsin⁡2xsin⁡xdx=∫0π2sin⁡xcos⁡xsin⁡x.dx=∫0π2cos⁡x.dx=[2sin⁡x]0π\Rightarrow {I_2} = \int_0^\pi {\dfrac{{\sin 2x}}{{\sin x}}dx} = \int_0^\pi {\dfrac{{2\sin x\cos x}}{{\sin x}}.dx} = \int_0^\pi {2\cos x.dx} = \left[ {2\sin x} \right]_0^\pi
⇒I2=[2sin⁡x]0π=2sin⁡π−2sin⁡0=0−0=0\Rightarrow {I_2} = \left[ {2\sin x} \right]_0^\pi = 2\sin \pi - 2\sin 0 = 0 - 0 = 0...................... (7)
Now from equation (5) substitute n = 1 we have,
⇒I1=I3\Rightarrow {I_1} = {I_3}
So from equation (6) we have,
⇒I1=I3=π\Rightarrow {I_1} = {I_3} = \pi
⇒I1=I3=I5=I7=......=π\Rightarrow {I_1} = {I_3} = {I_5} = {I_7} = ...... = \pi................. (8)
Now from equation (5) substitute n = 2 we have,
⇒I2=I4\Rightarrow {I_2} = {I_4}
So from equation (7) we have,
⇒I2=I4=0\Rightarrow {I_2} = {I_4} = 0
⇒I2=I4=I6=I8=......=0\Rightarrow {I_2} = {I_4} = {I_6} = {I_8} = ...... = 0................. (9)
Now check option (b) we have,
⇒∑m=110I2m+1=10π\Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = 10\pi
Consider the LHS we have,
⇒∑m=110I2m+1\Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m + 1}}}
Now expand the summation we have,
⇒∑m=110I2m+1=I3+I5+I7+I9+.....+I13\Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = {I_3} + {I_5} + {I_7} + {I_9} + ..... + {I_{13}}
Now from equation (8) we have,
⇒I1=I3=I5=I7=......=π\Rightarrow {I_1} = {I_3} = {I_5} = {I_7} = ...... = \pi
⇒∑m=110I2m+1=I3+I5+I7+I9+.....+I21=10π\Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = {I_3} + {I_5} + {I_7} + {I_9} + ..... + {I_{21}} = 10\pi
= LHS
So option (b) is also correct.
Now check option (c) we have,
(c)∑m=110I2m=0\left( c \right)\sum\limits_{m = 1}^{10} {{I_{2m}}} = 0
Consider the LHS we have,
⇒∑m=110I2m\Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m}}}
Now expand the summation we have,
⇒∑m=110I2m=I2+I4+I6+I8+.....+I20\Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m}}} = {I_2} + {I_4} + {I_6} + {I_8} + ..... + {I_{20}}
Now from equation (9) we have,
⇒I2=I4=I6=I8=......=0\Rightarrow {I_2} = {I_4} = {I_6} = {I_8} = ...... = 0
⇒∑m=110I2m=I2+I4+I6+I8+.....+I20=0\Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m}}} = {I_2} + {I_4} + {I_6} + {I_8} + ..... + {I_{20}} = 0
= LHS
So option (c) is also correct.
Hence options (a), (b) and (c) are the correct options.

Note: Whenever we face such types of questions the key concept we have to remember is the basic definite integral property so according to this property simplify the given integral as above then add them and again simplify using again different integral properties as above, then check the options one by one as above we will get the required answer.