Question

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

Answer 1

Hint: Here the very first thing that is to be considered is antiderivative is nothing but integration. That’s it! Now we will use the method of substitution such that we will put \[\sin x = u\] and then take the derivative. After that we will put the respective values in the integral. This will give the answer.

Complete step-by-step answer:
Given that find the antiderivative of \[\dfrac{{\cos x}}{{{{\sin }^2}x}}\]
Now let \[\sin x = u\]
Taking derivative on both the sides,
 \[\cos xdx = du\]
Thus,
 \[dx = \dfrac{{du}}{{\cos x}}\]
Now we will start our main solution,
 \[ = \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx} \]
Replacing the respective substitutions that is value of \[dx\] and \[\sin x\]
 \[ = \int {\dfrac{{\cos x}}{{{u^2}}} \times \dfrac{{du}}{{\cos x}}} \]
Cancelling \[\cos x\]
 \[ = \int {\dfrac{1}{{{u^2}}}du} \]
This can be written as,
 \[ = \int {{u^{ - 2}}du} \]
We know that \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \] applying this integral we get,
 \[ = \dfrac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}}\]
 \[ = \dfrac{{{u^{ - 1}}}}{{ - 1}}\]
On rewriting this equation,
 \[ = \dfrac{{ - 1}}{u}\]
Resubtitute the value of u.
 \[ = \dfrac{{ - 1}}{{\sin x}}\]
We know that \[\dfrac{1}{{\sin x}} = \cos ecx\]
Thus writing the final answer,
 \[ = - \cos ecx + C\]
This is the final answer.
 \[\int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx = - \cos ecx} \]
So, the correct answer is “- cosecx + C”.

Note: We can solve the same problem without substitution. In this method we will just shuffle the trigonometric functions. But the answer will remain the same.
 \[
  \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx} \\
   = \int {\dfrac{1}{{\sin x}}\dfrac{{\cos x}}{{\sin x}}dx} \\
   = \int {\cos ecx.\cot xdx} \\
   = - \cos ecx + C \\
\]
Because we know that \[\int {\cos ecx.\cot xdx} = - \cos ecx + C\] is the standard integration rule.
Note that integration and derivative are the reverse processes. Antiderivative is the same name for integration. This type of problem only involves the way we rotate or rearrange the trigonometric functions.