Let f(x) be a polynomial function of degree 2 satisfying ∫f(x)x3−1=ln∣∣x2+x+1x−1∣∣+2√3tan−1(2x+1√3)+c, where c is indefinite integration constant. Let ∫1−6cosecx6+f(sin x)d(sin x)=g(x)+k, where g(x) contains no constant term. Then limt→π2g(t) is equal to (where k is indefinite integration constant)

Let f(x) be a polynomial function of degree 2 satisfying ∫f(x)x3−1

1.	Let f(x) be a polynomial function of degree 2 satisfying ∫f(x)x3−1=ln∣∣x2+x+1x−1∣∣+2√3tan−1(2x+1√3)+c, where c is indefinite integration constant. Let ∫1−6cosecx6+f(sin x)d(sin x)=g(x)+k, where g(x) contains no constant term. Then limt→π2g(t) is equal to (where k is indefinite integration constant)
Answer» Let f(x) be a polynomial function of degree 2 satisfying $\int \frac{f (x)}{x^{3} - 1} = l n ∣ ∣ \frac{x^{2} + x + 1}{x - 1} ∣ ∣ + \frac{2}{\sqrt{3}} t a n^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + c,$ where c is indefinite integration constant. Let $\int \frac{1 - 6 c o s e c x}{6 + f (s i n x)} d (s i n x) = g (x) + k,$ where g(x) contains no constant term. Then $l i m t \to \frac{π}{2} g (t)$ is equal to (where k is indefinite integration constant)