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Question: How do you find the integral of \[\dfrac{1}{{{{\sin }^2}x}}\] ?...

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

Explanation

Solution

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⁡xu = \cot x . Because on further we can write cot⁡x=cos⁡xsin⁡x\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 1sin⁡2x\dfrac{1}{{{{\sin }^2}x}}
So that is ∫1sin⁡2xdx\int {\dfrac{1}{{{{\sin }^2}x}}dx}
Now we will substitute u=cot⁡xu = \cot x
On writing cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}}
Next we will differentiate the above terms with x.
du=ddxcos⁡xsin⁡xdu = \dfrac{d}{{dx}}\dfrac{{\cos x}}{{\sin x}}
With the help of product and quotient rule,
(uv)1=v.u1−u.v1v2{\left( {\dfrac{u}{v}} \right)^1} = \dfrac{{v.{u^1} - u.{v^1}}}{{{v^2}}}
du=−sin⁡x.sin⁡x−cos⁡x.cos⁡xsin⁡2xdxdu = \dfrac{{ - \sin x.\sin x - \cos x.\cos x}}{{{{\sin }^2}x}}dx
Taking the product of the terms,
du=−sin⁡2x−cos⁡2xsin⁡2xdxdu = \dfrac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}dx
Taking minus common from numerator terms,
du=−(sin⁡2x+cos⁡2x)sin⁡2xdxdu = \dfrac{{ - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x}}dx
We know the identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 then
du=−1sin⁡2xdxdu = \dfrac{{ - 1}}{{{{\sin }^2}x}}dx
−du=1sin⁡2xdx- du = \dfrac{1}{{{{\sin }^2}x}}dx
Now substituting the integral,
=∫−du= \int { - du}
Taking minus outside
=−∫du= - \int {du}
Taking integral,
=−u= - u
Substituting value of u,
=−cotx= - cotx
This is the answer.
So, the correct answer is “- cotx”.

Note : Note that this approach is used because no other approach is as easy as this. Also not that cot function is purposely used because only that function has the function and derivative are in numerator and denominator form. That helps to lead the problem in a smoother way. We can go for the ways like
1sin⁡2x=11−cos⁡2xdx\dfrac{1}{{{{\sin }^2}x}} = \dfrac{1}{{1 - {{\cos }^2}x}}dx
Or any other identity rearrangement but that will only lengthen the problem or can never lead to the answer. So do prefer this solution.