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  4. Let f (θ)= sin θ+∫ limits-π / 2π / 2( sin θ+ t cos θ) f ( t ) dt. Then the value of |∫ limits0π / 2 f(θ) d θ| is

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Q. Let $f (\theta)=\sin \theta+\int\limits_{-\pi / 2}^{\pi / 2}(\sin \theta+ t \cos \theta) f ( t ) dt$. Then the value of $\left|\int\limits_{0}^{\pi / 2} f(\theta) d \theta\right|$ is ___

JEE MainJEE Main 2022Integrals

Solution:

$f(\theta)=\sin \theta+\int\limits_{-\pi / 2}^{\pi / 2}(\sin \theta+t \cos \theta) f(t) d t $
$f(\theta)=\sin \theta+\sin \theta \int\limits_{-\pi / 2}^{\pi / 2} f(t) d t+\cos \theta \int\limits_{-\pi / 2}^{\pi / 2} t f(t) d t$
Let $A =\int\limits_{-\pi / 2}^{\pi / 2} f ( t ) dt , B =\int\limits_{-\pi / 2}^{\pi / 2} tf ( t ) dt$
$f(\theta)=\sin \theta+A \sin \theta+B \cos \theta$
$f(\theta)=(A+1) \sin \theta+B \cos \theta$
$A =\int\limits_{-\pi / 2}^{\pi / 2}( A +1) \sin t + B \cos t d t$
$A =2 B \ldots \ldots(1)$
$B =\int\limits_{-\pi / 2}^{\pi / 2} t (( A +1) \sin t + B \cos t )$
$B =\int\limits_{-\pi / 2}^{\pi / 2} t ( A +1) \sin t$
$B =( A +1) 2 \int\limits_{0}^{\pi / 2} t \sin t dt$
$B =( A +1) 2.1$
$2 A +2- B =0 \ldots \ldots(2)$
After solving
$B =-\frac{2}{3}, A =-\frac{4}{3}$
$\left|\int\limits_{0}^{\pi / 2} f(\theta) d \theta\right|=\left|\int\limits_{0}^{\pi / 2}-\frac{1}{3} \sin \theta-\frac{2}{3} \cos \theta\right|$
$=1$