Question

`\displaystyle\red{ \sf\lim_{x \to {0}^( + ) }\frac{1}{ {sin}^(2)x } \int _{ (x)/(2) }^(x) {sin}^( - 1) t \: dt}`​​

Answer 1

SOLUTION :-

• PLEASE CHECK THE ATTACHED FILE

Answer 2

Use L'Hopital's rule twice:

`\displaystyle \lim_(x\to0^+) \frac1{\sin^2(x)} \int_(\frac x2)^x \sin^(-1)(t) \, dt = \lim_(x\to0^+) \frac1{\sin^2(x)} \left(\int_0^x \sin^(-1)(t) \, dt - \int_0^(\frac x2) \sin^(-1)(t) \, dt\right) \\\\ = \lim_(x\to0^+) (\sin^(-1)(x) - \frac12 \sin^(-1)\left(\frac x2\right))/(2\sin(x)\cos(x)) ~~~~~~~~~~ (LHR) \\\\ = \lim_(x\to0^+) (2\sin^(-1)(x) - \sin^(-1)\left(\frac x2\right))/(2\sin(2x)) \\\\ = \lim_(x\to0^+) \frac{\frac2{√(1-x^2)} - \frac1{2\sqrt{1-\frac{x^2}4}}}{4\cos(2x)}  ~~~~~~~~~~ (LHR) \\\\ = \lim_(x\to0^+) \frac{\frac2{√(1-x^2)} - \frac1{√(4-x^2)}}{4\cos(2x)}`

The remaining limand is continuous at x = 0, and the limit is

`\displaystyle \lim_(x\to0^+) \frac1{\sin^2(x)} \int_(\frac x2)^x \sin^(-1)(t) \, dt = \frac{2-\frac12}4 = \boxed{\frac38}`