Question

If \[\left( {n - m} \right)\] is odd and \[\left| m \right| \ne \left| n \right|\], then what is the value of \[\int\limits_0^\pi {\cos mx\sin nxdx} \]?
A. \[\dfrac{{2n}}{{{n^2} - {m^2}}}\]
B. 0
C. \[\dfrac{{2n}}{{{m^2} - {n^2}}}\]
D. \[\dfrac{{2m}}{{{n^2} - {m^2}}}\]

Answer 1

Hint: Here, a definite integral is given. First, multiply and divide the given integral by 2. Then, simplify the term by applying the trigonometric formula \[2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)\]. After that, apply the integration rule \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\] and simplify the integral. Then, apply the integration formula \[\int\limits_a^b {\sin \left( {A + B} \right)xdx = \left[ { - \dfrac{{\cos \left( {A + B} \right)x}}{{\left( {A + B} \right)}}} \right]} _a^b\] and solve the integral. In the end, apply the upper and lower limits and simplify the terms to get the required answer.



Formula Used:\[2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)\]
Integration rule: \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\]
\[\int\limits_a^b {\sin \left( {A + B} \right)xdx = \left[ { - \dfrac{{\cos \left( {A + B} \right)x}}{{\left( {A + B} \right)}}} \right]} _a^b\]



Complete step by step solution:Given:
The definite integral is \[\int\limits_0^\pi {\cos mx\sin nxdx} \], where \[\left( {n - m} \right)\] is odd and \[\left| m \right| \ne \left| n \right|\].

Let consider,
\[I = \int\limits_0^\pi {\cos mx\sin nxdx} \]
Multiply and divide the right-hand side by 2.
\[I = \dfrac{1}{2}\int\limits_0^\pi {2\cos mx\sin nxdx} \]
Now use the trigonometric identity \[2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)\].
\[I = \dfrac{1}{2}\int\limits_0^\pi {\left[ {\sin \left( {mx + nx} \right) - \sin \left( {mx - nx} \right)} \right]dx} \]
\[ \Rightarrow I = \dfrac{1}{2}\int\limits_0^\pi {\left[ {\sin \left( {m + n} \right)x - \sin \left( {m - n} \right)x} \right]dx} \]
Apply the integration rule \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\].
\[I = \dfrac{1}{2}\left[ {\int\limits_0^\pi {\sin \left( {m + n} \right)xdx - \int\limits_0^\pi {\sin \left( {m - n} \right)xdx} } } \right]\]
Now solve the integrals by applying the integration formula \[\int\limits_a^b {\sin \left( {A + B} \right)xdx = \left[ { - \dfrac{{\cos \left( {A + B} \right)x}}{{\left( {A + B} \right)}}} \right]} _a^b\].
We get,
 \[I = \dfrac{1}{2}\left[ {\left[ { - \dfrac{{\cos \left( {m + n} \right)x}}{{\left( {m + n} \right)}}} \right]_0^\pi - \left[ { - \dfrac{{\cos \left( {m - n} \right)x}}{{\left( {m - n} \right)}}} \right]_0^\pi } \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\left[ { - \dfrac{{\cos \left( {m + n} \right)x}}{{\left( {m + n} \right)}}} \right]_0^\pi + \left[ {\dfrac{{\cos \left( {m - n} \right)x}}{{\left( {m - n} \right)}}} \right]_0^\pi } \right]\]
Apply the upper and lower limits.
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\left[ { - \dfrac{{\cos \left( {m + n} \right)\pi }}{{\left( {m + n} \right)}} + \dfrac{{\cos \left( {\left( {m + n} \right)0} \right)}}{{\left( {m + n} \right)}}} \right] + \left[ {\dfrac{{\cos \left( {m - n} \right)\pi }}{{\left( {m - n} \right)}} - \dfrac{{\cos \left( {\left( {m - n} \right)0} \right)}}{{\left( {m - n} \right)}}} \right]} \right]\]
Use the trigonometric properties \[\cos 0 = 1\] and \[\cos n\pi = - 1\].
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\left[ { - \dfrac{{ - 1}}{{\left( {m + n} \right)}} + \dfrac{1}{{\left( {m + n} \right)}}} \right] + \left[ {\dfrac{{ - 1}}{{\left( {m - n} \right)}} - \dfrac{1}{{\left( {m - n} \right)}}} \right]} \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\left[ {\dfrac{1}{{m + n}} + \dfrac{1}{{m + n}}} \right] + \left[ {\dfrac{{ - 1}}{{m - n}} - \dfrac{1}{{m - n}}} \right]} \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\dfrac{2}{{m + n}} - \dfrac{2}{{m - n}}} \right]\]
\[ \Rightarrow I = \dfrac{1}{{m + n}} - \dfrac{1}{{m - n}}\]
Solve the right-hand side.
\[ \Rightarrow I = \dfrac{{\left( {m - n} \right) - \left( {m + n} \right)}}{{\left( {m + n} \right)\left( {m - n} \right)}}\]
\[ \Rightarrow I = \dfrac{{m - n - m - n}}{{{m^2} - {n^2}}}\]
\[ \Rightarrow I = \dfrac{{ - 2n}}{{{m^2} - {n^2}}}\]
\[ \Rightarrow I = \dfrac{{2n}}{{{n^2} - {m^2}}}\]
Therefore, \[\int\limits_0^\pi {\cos mx\sin nxdx} = \dfrac{{2n}}{{{n^2} - {m^2}}}\].



Option ‘A’ is correct


Note: Students get confused and make mistake while integrating \[\int\limits_a^b {\sin \left( {m + n} \right)xdx} \]. They apply the formula \[\int\limits_a^b {\sin \left( {m + n} \right)xdx} = \left[ { - \cos \left( {m + n} \right)x} \right]_a^b\], which is an incorrect formula. The correct formula is \[\int\limits_a^b {\sin \left( {m + n} \right)xdx} = \left[ {\dfrac{{ - \cos \left( {m + n} \right)x}}{{m + n}}} \right]_a^b\].