Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 14th Jul 2024
Total views: 380.4k
Views today: 7.80k
Answer
VerifiedVerified
380.4k+ views
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,
 \[ = - cotx\]
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
 \[\dfrac{1}{{{{\sin }^2}x}} = \dfrac{1}{{1 - {{\cos }^2}x}}dx\]
Or any other identity rearrangement but that will only lengthen the problem or can never lead to the answer. So do prefer this solution.