# Choose the correct answer in $\Large \int \normalsize\frac{dx}{\sin^2x\cos^2x}$ equals

$\begin{array}{1 1} (A)\;\tan x+\cot x+C & (B)\;\tan x-\cot x+C\\(C)\;\tan x\cot x+C & (D)\;\tan x-\cot2x+C\end{array}$

Toolbox:
• (i)Method of substitution :
• Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
• Consider $I=\int f(x)dx.$
• Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
• $\Rightarrow$dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{dx}{\sin^2x\cos^2x}dx$.

We know $\sin^2x+\cos^2x=1.$

Hence we can write I as,

$I=\int \frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}dx.$

Now separating the terms we get,

$I=\int \big(\frac{1}{\cos^2x}+\frac{1}{\sin^2x}\big)dx$.

But $\frac{1}{\cos^2x}=\sec^2x$ and $\frac{1}{\sin^2x}=cosec^2x.$

$\;\;=\int\sec^2xdx+\int cosec^2xdx.$
On integrating we get,

$\tan x-cot x+c.$

Hence the correct answer is B.