Question

Solve the following integral:
 $$\int \dfrac{1}{\cos^{2} x\left( 1-\tan x\right)^{2} } dx$$.

Answer 1

Hint:In this question it is given that we have to solve $$\int \dfrac{1}{\cos^{2} x\left( 1-\tan x\right)^{2} } dx$$.To find the solution we have to use the substitution method, i.e, if derivative of any part of denominator is there in numerator then we have to substitute this part of denominator as a different variable and after by using suitable integration formula we will get our solution.
Complete step by step answer:
Given integration,
$$\int \dfrac{1}{\cos^{2} x\left( 1-\tan x\right)^{2} } dx$$
=$$\int \dfrac{\sec^{2} x}{\left( 1-\tan x\right)^{2} } dx$$ [$$\because \dfrac{1}{\cos x} =\sec x$$]
Now we are going to use, substitution method.
Let $$1-\tan x=t$$
Differentiating both side, we get,
$$-\sec^{2} x\ dx=dt$$
$$\Rightarrow \sec^{2} x\ dx=-dt$$
So by substituting, the above integration can be written as,
$$\int \dfrac{-dt}{t^{2}}$$
=$$-\int \dfrac{dt}{t^{2}}$$
=$$-\int t^{-2}dt$$
Now by using $$\dfrac{d}{dx} \left( x^{n}\right) =\dfrac{x^{n+1}}{n+1} +c$$, we get,
=$$-\left( \dfrac{t^{-2+1}}{-2+1} \right) +c$$ [where c is integration constant]
=$$-\left( \dfrac{t^{-1}}{-1} \right) +c$$
=$$\dfrac{1}{t} +c$$
Now by putting the value of t, we get,
$$\int \dfrac{1}{\cos^{2} x\left( 1-\tan x\right)^{2} } dx$$=$$\dfrac{1}{\left( 1-\tan x\right) } +c$$.
Which is our required solution.
Note: To solve this type of question you have to keep in mind that if the derivative of any term in the denominator is there in the numerator then we have to use the substitution method, like we have used in the above solution, i.e, derivative of $$1-\tan x$$ was there in the numerator which is $$\sec^{2} x$$.