Question

Let \[{I_1} = \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \], \[{I_2} = \int_a^{\pi - a} {f\left( {\sin x} \right)dx} \], then find the value of \[{I_2}\] in term of \[{I_1}\].
A. \[\dfrac{\pi }{2}{I_1}\]
B. \[\pi {I_1}\]
C. \[\dfrac{2}{\pi }{I_1}\]
D. \[2{I_1}\]

Answer 1

Hint: To solve the given question we will apply definite integral property. First we will apply the property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \] in \[{I_1}\] and simplify it by using trigonometry supplementary formula. Then substitute \[\int_a^{\pi - a} {xf\left( {\sin x} \right)dx} = {I_1}\] and \[\int_a^{\pi - a} {f\left( {\sin x} \right)dx} = {I_2}\] and calculate the value of \[{I_2}\] in term \[{I_1}\].



Formula Used:Definite integral property:
\[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]
Trigonometry supplementary formula:
\[\sin \left( {\pi - \theta } \right) = \sin \theta \]



Complete step by step solution:Given integration is \[{I_1} = \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Now applying the definite integral property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - a + a - x} \right)f\left( {\sin \left( {\pi - a + a - x} \right)} \right)dx} \]
Simplify the above equation:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - x} \right)f\left( {\sin \left( {\pi - x} \right)} \right)dx} \]
Now we will apply trigonometry supplementary formula \[\sin \left( {\pi - \theta } \right) = \sin \theta \]:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\left( {\pi - x} \right)f\left( {\sin x} \right)dx} \]
Now break the integration:
\[ \Rightarrow {I_1} = \int_a^{\pi - a} {\pi f\left( {\sin x} \right)dx} - \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Rewrite the above equation:
\[ \Rightarrow {I_1} = \pi \int_a^{\pi - a} {f\left( {\sin x} \right)dx} - \int_a^{\pi - a} {xf\left( {\sin x} \right)dx} \]
Substitute \[\int_a^{\pi - a} {xf\left( {\sin x} \right)dx} = {I_1}\] and \[\int_a^{\pi - a} {f\left( {\sin x} \right)dx} = {I_2}\] in the above equation:
\[ \Rightarrow {I_1} = \pi {I_2} - {I_1}\]
Add \[{I_1}\] on both sides of the equation:
\[ \Rightarrow {I_1} + {I_1} = \pi {I_2} - {I_1} + {I_1}\]
\[ \Rightarrow 2{I_1} = \pi {I_2}\]
Divide both sides by \[\pi \]:
\[ \Rightarrow \dfrac{2}{\pi }{I_1} = {I_2}\]



Option ‘C’ is correct



Note: Students often mistake we they apply trigonometry supplementary formula. They confused with complementary formula. They applied \[\sin \left( {\pi - \theta } \right) = - \cos \theta \] which incorrect formula. The correct formula is \[\sin \left( {\pi - \theta } \right) = \sin \theta \].