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# How do you find the integral of $\dfrac{1}{{{{\sin }^2}x}}$ ?

Last updated date: 14th Jul 2024
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Hint: Here we have to find the integral of a given function. We will use the trigonometric ratios and try to write the given ratio in standard integral form. Here first we will use the substitution method. Let’s put $u = \cot x$ . Because on further we can write $\cot x = \dfrac{{\cos x}}{{\sin x}}$ on taking differentiation with the help of product and quotient rule will help in solving the problem in lesser steps and the best approach.

Complete step by step solution:
Given that find the integral of $\dfrac{1}{{{{\sin }^2}x}}$
So that is $\int {\dfrac{1}{{{{\sin }^2}x}}dx}$
Now we will substitute $u = \cot x$
On writing $\cot x = \dfrac{{\cos x}}{{\sin x}}$
Next we will differentiate the above terms with x.
$du = \dfrac{d}{{dx}}\dfrac{{\cos x}}{{\sin x}}$
With the help of product and quotient rule,
${\left( {\dfrac{u}{v}} \right)^1} = \dfrac{{v.{u^1} - u.{v^1}}}{{{v^2}}}$
$du = \dfrac{{ - \sin x.\sin x - \cos x.\cos x}}{{{{\sin }^2}x}}dx$
Taking the product of the terms,
$du = \dfrac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}dx$
Taking minus common from numerator terms,
$du = \dfrac{{ - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x}}dx$
We know the identity ${\sin ^2}x + {\cos ^2}x = 1$ then
$du = \dfrac{{ - 1}}{{{{\sin }^2}x}}dx$
$- du = \dfrac{1}{{{{\sin }^2}x}}dx$
Now substituting the integral,
$= \int { - du}$
Taking minus outside
$= - \int {du}$
Taking integral,
$= - u$
Substituting value of u,
$= - cotx$
$\dfrac{1}{{{{\sin }^2}x}} = \dfrac{1}{{1 - {{\cos }^2}x}}dx$