\(∫\frac {sin^2 x-cos^2 x}{sin^2 x cos^2 x }\ dx \ is\ equal \ to\)
\(tan\ x + cot\ x +C\)
\(tan \ x + cosec \ x +C\)
\(-tan\ x + cot\ x +C\)
\(tan \ x + sec\ x +C\)
\(∫\frac {sin^2 x-cos^2 x}{sin^2 x cos^2 x }\ dx\)
= \(∫(\frac {sin^2 x}{sin^2 x cos^2 x }\ dx\) - \(∫\frac {cos^2 x}{sin^2 x cos^2 x })\ dx\)
= \(∫(sec^2 x - cosec^2 x) dx\)
= \(tan\ x + cot \ x +C\)
Hence, the correct Answer is (A): \(tan\ x + cot \ x +C\)
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