Let \(f(\theta)=\sin \theta+\int_{-\pi / 2}^{\pi / 2}(\sin \theta+t \cos \theta) f(t) d t\). Then the value of \(\left|\int_{0}^{\pi / 2} f(\theta) d \theta\right|\) is ______.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Definite Integrals

Let

f (θ) = sin θ + \int_{- π /2}^{π /2} (sin θ + t cos θ) f (t) d t

. Then the value of

\int_{0}^{π /2} f (θ) d θ

is ______.

Answer: 1

Write

f (θ) = sin θ + sin θ \int_{- π /2}^{π /2} f (t) d t + cos θ \int_{- π /2}^{π /2} t f (t) d t

.

Let

A = \int_{- π /2}^{π /2} f (t) d t

and

B = \int_{- π /2}^{π /2} t f (t) d t

. Then

f (θ) = (A + 1) sin θ + B cos θ .

Substitute back:

A = \int_{- π /2}^{π /2} [(A + 1) sin t + B cos t] d t = 2 B .

B = \int_{- π /2}^{π /2} t [(A + 1) sin t + B cos t] d t = (A + 1) \int_{- π /2}^{π /2} t sin t d t = 2 (A + 1) .

From

A = 2 B

and

B = 2 (A + 1)

A = 4 (A + 1) ⟹ - 3 A = 4 ⟹ A = - \frac{4}{3}, B = - \frac{2}{3} .

Thus

f (θ) = - \frac{1}{3} sin θ - \frac{2}{3} cos θ

\int_{0}^{π /2} f (θ) d θ = \frac{1}{3} cos θ_{0}^{π /2} - \frac{2}{3} sin θ_{0}^{π /2} = - \frac{1}{3} - \frac{2}{3} = 1.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer