Question

If we have an integral as \[{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx,n = 0,1,2,3.....} \] then
$\left( a \right){I_n} = {I_{n + 2}}$
$\left( b \right)\sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = 10\pi $
$\left( c \right)\sum\limits_{m = 1}^{10} {{I_{2m}}} = 0$
$\left( d \right){I_n} = {I_{n + 1}}$

Answer 1

Hint: In this particular question use the property of definite integral i.e. $\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:
\[{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx} \]....................... (1)
Now according to the definite integral property $\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} $ we have,
\[ \Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin n\left( { - \pi + \pi - x} \right)}}{{\left( {1 + {\pi ^{\left( { - \pi + \pi - x} \right)}}} \right)\sin \left( { - \pi + \pi - x} \right)}}dx} \]
Now simplify this using the property that (sin (-x) = -sin x) so we have,
\[ \Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin n\left( { - x} \right)}}{{\left( {1 + {\pi ^{\left( { - x} \right)}}} \right)\sin \left( { - x} \right)}}dx} \]
\[ \Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{ - \sin nx}}{{ - \left( {1 + {\pi ^{\left( { - x} \right)}}} \right)\sin x}}dx} \]
\[ \Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + \dfrac{1}{{{\pi ^x}}}} \right)\sin x}}dx} \]
\[ \Rightarrow {I_n} = \int_{ - \pi }^\pi {\dfrac{{{\pi ^x}\sin nx}}{{\left( {{\pi ^x} + 1} \right)\sin x}}dx} \]....................... (2)
Now add equation (1) and (2) we have,
\[ \Rightarrow 2{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx} + \int_{ - \pi }^\pi {\dfrac{{{\pi ^x}\sin nx}}{{\left( {{\pi ^x} + 1} \right)\sin x}}dx} \]
\[ \Rightarrow 2{I_n} = \int_{ - \pi }^\pi {\dfrac{{\left( {1 + {\pi ^x}} \right)\sin nx}}{{\left( {1 + {\pi ^x}} \right)\sin x}}dx} \]
\[ \Rightarrow 2{I_n} = \int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\sin x}}dx} \]
\[ \Rightarrow {I_n} = \dfrac{1}{2}\int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\sin x}}dx} \]
Now as we know that $\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 $\dfrac{{\sin nx}}{{\sin x}} = $even function.
\[ \Rightarrow {I_n} = \dfrac{1}{2}\int_{ - \pi }^\pi {\dfrac{{\sin nx}}{{\sin x}}dx} = \dfrac{2}{2}\int_0^\pi {\dfrac{{\sin nx}}{{\sin x}}dx} \]
\[ \Rightarrow {I_n} = \int_0^\pi {\dfrac{{\sin nx}}{{\sin x}}dx} \]................ (3)
Now substitute in place of n, (n + 2) so we have,
\[ \Rightarrow {I_{n + 2}} = \int_0^\pi {\dfrac{{\sin \left( {n + 2} \right)x}}{{\sin x}}dx} \]................ (4)
Now subtract equation (3) from equation (4) we have,
\[ \Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {\dfrac{{\sin \left( {n + 2} \right)x - \sin nx}}{{\sin x}}dx} \]
Now as we know that sin C – sin D = $2\sin \left( {\dfrac{{C - D}}{2}} \right)\cos \left( {\dfrac{{C + D}}{2}} \right)$ so use this property we have,
\[ \Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {\dfrac{{2\sin \left( {\dfrac{{nx + 2x - nx}}{2}} \right)\cos \left( {\dfrac{{nx + 2x + nx}}{2}} \right)}}{{\sin x}}dx} \]
\[ \Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {\dfrac{{2\sin x\cos \left( {n + 1} \right)x}}{{\sin x}}dx} \]
\[ \Rightarrow {I_{n + 2}} - {I_n} = \int_0^\pi {2\cos \left( {n + 1} \right)xdx} \]
Now integrate it we have,
\[ \Rightarrow {I_{n + 2}} - {I_n} = \left[ {\dfrac{{2\sin \left( {n + 1} \right)x}}{{n + 1}}} \right]_0^\pi \]
Now apply integrating limits we have,
\[ \Rightarrow {I_{n + 2}} - {I_n} = \left[ {\dfrac{{2\sin \left( {n + 1} \right)\pi }}{{n + 1}} - \dfrac{{2\sin 0}}{{n + 1}}} \right]\]
Now as it is given that n = 0, 1, 2, 3.... so n is an integer so, \[\sin \left( {n + 1} \right)\pi = 0\] and sin 0 = 0 so we h ave,
\[ \Rightarrow {I_{n + 2}} - {I_n} = \left[ {0 - 0} \right] = 0\]
\[ \Rightarrow {I_n} = {I_{n + 2}}\]............... (5)
So option (a) is correct.
Now in equation (3) substitute n = 1 we have,
\[ \Rightarrow {I_1} = \int_0^\pi {\dfrac{{\sin x}}{{\sin x}}dx} = \int_0^\pi {1.dx} = \left[ x \right]_0^\pi = \pi \]..................... (6)
Now in equation (3) substitute n = 2 we have,
\[ \Rightarrow {I_2} = \int_0^\pi {\dfrac{{\sin 2x}}{{\sin x}}dx} = \int_0^\pi {\dfrac{{2\sin x\cos x}}{{\sin x}}.dx} = \int_0^\pi {2\cos x.dx} = \left[ {2\sin x} \right]_0^\pi \]
\[ \Rightarrow {I_2} = \left[ {2\sin x} \right]_0^\pi = 2\sin \pi - 2\sin 0 = 0 - 0 = 0\]...................... (7)
Now from equation (5) substitute n = 1 we have,
\[ \Rightarrow {I_1} = {I_3}\]
So from equation (6) we have,
\[ \Rightarrow {I_1} = {I_3} = \pi \]
\[ \Rightarrow {I_1} = {I_3} = {I_5} = {I_7} = ...... = \pi \]................. (8)
Now from equation (5) substitute n = 2 we have,
\[ \Rightarrow {I_2} = {I_4}\]
So from equation (7) we have,
\[ \Rightarrow {I_2} = {I_4} = 0\]
\[ \Rightarrow {I_2} = {I_4} = {I_6} = {I_8} = ...... = 0\]................. (9)
Now check option (b) we have,
$ \Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = 10\pi $
Consider the LHS we have,
$ \Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m + 1}}} $
Now expand the summation we have,
$ \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,
\[ \Rightarrow {I_1} = {I_3} = {I_5} = {I_7} = ...... = \pi \]
$ \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,
$\left( c \right)\sum\limits_{m = 1}^{10} {{I_{2m}}} = 0$
Consider the LHS we have,
$ \Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m}}} $
Now expand the summation we have,
$ \Rightarrow \sum\limits_{m = 1}^{10} {{I_{2m}}} = {I_2} + {I_4} + {I_6} + {I_8} + ..... + {I_{20}}$
Now from equation (9) we have,
\[ \Rightarrow {I_2} = {I_4} = {I_6} = {I_8} = ...... = 0\]
$ \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.