Question

Let f(x) be a polynomial function of degree 2 satisfying $\int \frac{f\left(x\right)}{x^3-1}=ln\left|\frac{x^2+x+1}{x-1}\right|+\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+c,$ where c is indefinite integration constant.
Let $\int \frac{1-6cosecx}{6+f\left(sinx\right)}d\left(sinx\right)=g\left(x\right)+k,$ where g(x) contains no constant term. Then $\underset{t\to \frac{\pi }{2}}{lim}g\left(t\right)$ is equal to (where k is indefinite integration constant)

• A. $ln1$
• B. $ln2$
• C. $ln3$
• D. $ln4$

Answer 1

The correct option is C $ln3$

$\int \frac{f\left(x\right)}{x^3-1}dx=ln\left|\frac{x^2+x+1}{x-1}\right|+\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+3$
Differentiating both sides, we get
$\frac{f\left(x\right)}{x^3-1}=\frac{x-1}{x^2+x+1}\frac{(x-1(2x+1)-(x^2+x+1\left).1\right)}{(x-1{)}^2}+\frac{2}{\sqrt{3}}.\frac{1}{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}.\frac{2}{\sqrt{3}}=\frac{x^2-2x-2}{x^3-1}+\frac{1}{x^2+x+1}f\left(x\right)=x^2-x-3Now.I=\int \frac{1-6cosecx}{6+si{n}^{2}x-sinx-3}d\left(sinx\right)Putsinx=t=\int \frac{1-\frac{6}{t}}{t^2-t+3}dt=\int \frac{\frac{1}{t^2}-\frac{6}{t^3}}{1-\frac{1}{t}+\frac{3}{t^2}}dtLet1-\frac{1}{t}+\frac{3}{t^2}=P=ln(1-\frac{1}{t}+\frac{3}{t^2}=)+k=ln(1-\frac{1}{sinx}+\frac{3}{si{n}^{2}x})+k$
$g\left(t\right)=ln(1-\frac{1}{sint+\frac{3}{si{n}^{2}t}})\underset{t\to \frac{\pi }{2}}{lim}g\left(t\right)=ln3$