The value of the integral $\frac{48}{\pi^4} \int_0^\pi \left(\frac{3\pi x^2}{2} - x^3\right) \frac{\sin x}{1+\cos^2 x} dx$ is equal to ____.

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Definite Integrals

The value of the integral

\frac{48}{π ^{4}} \int_{0}^{π} (\frac{3 π x ^{2}}{2} - x^{3}) \frac{s i n x}{1 + c o s ^{2} x} d x

is equal to ____.

Answer: 6

Let

I = \frac{48}{π ^{4}} \int_{0}^{π} x^{2} (\frac{3 π}{2} - x) \frac{s i n x}{1 + c o s ^{2} x} d x

.

Applying the King property

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

repeatedly and adding equations, we can simplify the integrand. After multiple applications, we obtain:

I = \frac{12}{π} \int_{0}^{π} \frac{sin x}{1 + cos ^{2} x} d x

Let

cos x = t

, then

- sin x d x = d t

. When

x = 0

t = 1

; when

x = π

t = - 1

I = \frac{12}{π} \int_{1}^{- 1} \frac{- d t}{1 + t ^{2}} = \frac{12}{π} \int_{- 1}^{1} \frac{d t}{1 + t ^{2}}

= \frac{12}{π} [arctan t]_{- 1}^{1} = \frac{12}{π} (\frac{π}{4} - (- \frac{π}{4})) = \frac{12}{π} \cdot \frac{π}{2} = 6

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer