Let $f$ be integrable over $\left[ {0,\,\,a} \right]$ for any real $a$. If we define ${I_1} = \mathop \smallint \nolimits_0^{\pi /2} \,\,\cos \,\,{\rm{\theta }}\,\,f(\sin \,\,{\rm{\theta }}\,\, + \,\,{\cos ^2}\,\,{\rm{\theta }})\,\,\;d{\rm{\theta }}$ and ${I_2} = \mathop \smallint \nolimits_0^{\pi /2} \sin \,\,2{\rm{\theta }}\,\,f(\sin \,\,{\rm{\theta }}\,\, + \,\,{\cos ^2}\,\,{\rm{\theta }})\;\,\,d{\rm{\theta }}$, then

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

Let

f

be integrable over

\left[ {0,\,\,a} \right]

for any real

a

. If we define

{I_1} = \int \nolimits_0^{\pi /2} \,\,\cos \,\,\mathrm{\theta }\,\,f(\sin \,\,\mathrm{\theta }\,\, + \,\,{\cos ^2}\,\,\mathrm{\theta })\,\,\;d\mathrm{\theta }

and

{I_2} = \int \nolimits_0^{\pi /2} \sin \,\,2\mathrm{\theta }\,\,f(\sin \,\,\mathrm{\theta }\,\, + \,\,{\cos ^2}\,\,\mathrm{\theta })\;\,\,d\mathrm{\theta }

, then

View answer and explanation

Correct option: A

{I_1} = {I_2}

Option A is correct because by substituting sin(

\theta

) + cos²(

\theta

) = t, the integral I1 - I2 becomes zero, implying I1 = I2.