Let ${I_1} = \mathop \smallint \nolimits_a^{\pi - a} \,\,xf\,\,(\sin \,\,x)\;\,\,dx,\;\,\,{I_2} = \mathop \smallint \nolimits_a^{\pi - a} \,\,f\;(\sin \,\,x)\;\,\,dx$, then ${I_2}$ is equal to

$\frac{2}{{\rm{\pi }}}\,\,{I_1}$

$\frac{{\rm{\pi }}}{2}\,\,{I_1}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

Let

{I_1} = \int \nolimits_a^{\pi - a} \,\,xf\,\,(\sin \,\,x)\;\,\,dx,\;\,\,{I_2} = \int \nolimits_a^{\pi - a} \,\,f\;(\sin \,\,x)\;\,\,dx

, then

{I_2}

is equal to

View answer and explanation

Correct option: C

\frac{2}{\mathrm{\pi }}\,\,{I_1}

Option C (

\frac{2}{\mathrm{\pi }}\I_1

) is correct because it is derived by using the property of definite integrals and solving for I2 in terms of I1.

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