Question

Find the integral $\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$?
(a) $\tan x-\cot x+c$
(b) $\tan x-x+1$
(c) $\tan x-x$
(d) $\tan x+x$

Answer 1

Assume the given integral as ‘I’. Use the trigonometric identity $2\sin x\cos x=\sin 2x$ and simplify the function inside the integral. Now, Use the conversion $\dfrac{1}{{{\sin }^{2}}x}=\cos e{{c}^{2}}x$ and integrate the cosecant function using the formula $\int{\cos e{{c}^{2}}\left( ax+b \right)dx}=-\dfrac{1}{a}\cot \left( ax+b \right)$. Further, simplify the expression by using the formula $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ and the conversions $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{\cos x}{\sin x}=\cot x$to get the answer.

Complete step by step answer:
Here we have been asked to integrate the function $\dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}$. Let us assume the integral as I so we have,
$\Rightarrow I=\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$
We can write the above integral as:
$\Rightarrow I=\int{\dfrac{dx}{{{\left( \sin x\cos x \right)}^{2}}}}$
Using the trigonometric identity $2\sin x\cos x=\sin 2x$ we get,
$\begin{aligned} & \Rightarrow I=\int{\dfrac{dx}{{{\left( \dfrac{\sin 2x}{2} \right)}^{2}}}} \\\ & \Rightarrow I=\int{\dfrac{4dx}{{{\sin }^{2}}2x}} \\\ \end{aligned}$
Since 4 is a constant so it can be taken out of the integral and using the conversion $\dfrac{1}{{{\sin }^{2}}x}=\cos e{{c}^{2}}x$ we have,
$\Rightarrow I=4\int{\cos e{{c}^{2}}2xdx}$
Now, using the integration formula of the co – secant function given as $\int{\cos e{{c}^{2}}\left( ax+b \right)dx}=-\dfrac{1}{a}\cot \left( ax+b \right)$ where a and b are constants we get,
$\begin{aligned} & \Rightarrow I=4\left( -\dfrac{1}{2}\cot 2x \right) \\\ & \Rightarrow I=-2\left( \cot 2x \right) \\\ \end{aligned}$
Using the conversion $\cot x=\dfrac{\cos x}{\sin x}$ we get,
$\Rightarrow I=-2\left( \dfrac{\cos 2x}{\sin 2x} \right)$
Further using the trigonometric identity $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ in the numerator and $2\sin x\cos x=\sin 2x$ in the denominator we get,
$\begin{aligned} & \Rightarrow I=-2\left( \dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{2\sin x\cos x} \right) \\\ & \Rightarrow I=\left( \dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x\cos x} \right) \\\ \end{aligned}$
Breaking the terms we get,
$\Rightarrow I=\dfrac{{{\sin }^{2}}x}{\sin x\cos x}-\dfrac{{{\cos }^{2}}x}{\sin x\cos x}$
Cancelling the common terms and using the conversions $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{\cos x}{\sin x}=\cot x$ we get,
$\begin{aligned} & \Rightarrow I=\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x} \\\ & \Rightarrow I=\tan x-\cot x \\\ \end{aligned}$
Now, since the given integral is an indefinite integral and therefore we need to add a constant of integration (c) in the expression obtained for I. So we get,
$\therefore I=\tan x-\cot x+c$

So, the correct answer is “Option a”.

Note: Remember all the trigonometric identities and the integral and differential formulas of basic functions such as trigonometric function, logarithmic function, inverse trigonometric functions etc. At last, do not forget to add the constant of integration (c) as we are finding indefinite integral and not definite integral. If options are given, you can also find the correct answer by differentiating the functions one by one. The option which will give the function present inside the integral sign will be our answer.