Let sin-1 x = t
∴ \(\frac{1}{\sqrt {1-x^2}}dx=dt\)
\(\Rightarrow \int\frac{\sin^{-1}x}{\sqrt {1-x^2}}dx=\int t dt\)
= \(\frac{t^2}{2}+C\)
= \(\frac{(sin^{-1} x)^2}{2}+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males. A person is selected at random from these 50 people and tested for lung related problems. Based on the given information answer the following questions: 
(i) What is the probability that selected person is a female?
(ii) If a male person is selected, what is the probability that he will not be suffering from lung problems?
(iii)(a) A person selected at random is detected with lung complications. Find the probability that selected person is a female.
OR
(iii)(b) A person selected at random is not having lung problems. Find the probability that the person is a male.
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C