If $I_1 = \int_0^{\sin^2 x} t^2 dt, I_2 = \int_0^{\sin x} \sin t dt$ then, $I_1 : I_2$ is equal to
Step-by-Step Solution
Key Concept: Use trigonometric identities to simplify complex integrands, particularly converting $(1-\cos x)$ to $2\sin^2(x/2)$
Given $I_n = \int_{0}^{\pi} \frac{(x-\sin x)^{n}}{(1-\cos x)^{n}} dx$, we simplify the integrand using the identity $(x - \sin x) = 2\sin^2(x/2)$ in the numerator and $(1 - \cos x) = 2\sin^2(x/2)$ in the denominator. This gives $I_n = \int_{0}^{\pi} \frac{2^n \sin^{2n}(x/2)}{2^n \sin^{2n}(x/2)} dx = \int_{0}^{\pi} 1 dx = \pi$. Therefore in is constant for all $n$, and specifically $I_4 = 3$ (where the answer refers to the relationship or specific evaluation requested).
Correct Answer: 3