Question

Find the value of $ \int {\dfrac{{2\cos x}}{{\left( {1 - \sin x} \right)\left( {1 + {{\sin }^2}x} \right)}}} dx $ .

Answer 1

Hint: The given question is related to the concept of integration. The sub-topic of integration is limited. Here in this given question, we will have to solve the limit. The limits can be solved using the methods of direct substitution, factoring and cancelling, expanding and simplifying, combining factors, by multiplying by the conjugate and many more. We will solve this question by using a partial fraction method.
For rational function $ \dfrac{{p{x^2} + qx + r}}{{\left( {x - a} \right)\left( {{x^2} + bx + c} \right)}} $ , the form of partial fraction is $ \dfrac{A}{{x - a}} + \dfrac{{Bx + C}}{{{x^2} + bx + c}} $ , where A and B are real numbers.

Complete step-by-step answer:
Given is $ \int {\dfrac{{2\cos x}}{{\left( {1 - \sin x} \right)\left( {1 + {{\sin }^2}x} \right)}}} dx $
Let,
 $
   \Rightarrow \sin x = y \\
   \Rightarrow \cos x.dx = dy \\
  $
Substituting these values in the given limit, we get
 $ \therefore \int {\dfrac{{2\cos x}}{{\left( {1 - \sin x} \right)\left( {1 + {{\sin }^2}x} \right)}}.} dx = \int {\dfrac{2}{{\left( {1 - y} \right)\left( {1 + {y^2}} \right)}}} .dy $
Now, by partial fraction, let
 $ \Rightarrow \dfrac{2}{{\left( {1 - y} \right)\left( {1 + {y^2}} \right)}} = \dfrac{A}{{1 - y}} + \dfrac{{By + C}}{{1 + {y^2}}} $
Where real numbers A and B are to be determined.
Solving it gives,
 $
   \Rightarrow 2 = A\left( {1 + {y^2}} \right) + \left( {By + C} \right)\left( {1 - y} \right) \\
   \Rightarrow 2 = A + A{y^2} + By + C - B{y^2} - Cy \\
   \Rightarrow 2 = \left( {A - B} \right){y^2} + \left( {B - C} \right)y + A + C \\
  $
Now, we compare the coefficients of $ {y^2},y $ and the constant terms. We get,
 $
   \Rightarrow A - B = 0 \\
   \Rightarrow B - C = 0 \\
   \Rightarrow A + C = 2 \;
  $
After solving these equations, we get $ A = B = C = 1 $
 $
   \Rightarrow \int {\dfrac{{2\cos x}}{{\left( {1 - \sin x} \right)\left( {1 + {{\sin }^2}x} \right)}}} .dx \\
   = \int {\dfrac{2}{{\left( {1 - y} \right)\left( {1 + {y^2}} \right)}}} .dy \\
   = \int {\dfrac{1}{{1 - y}}} .dy + \int {\dfrac{{y + 1}}{{1 + {y^2}}}} .dt \\
   = \int {\dfrac{1}{{1 - y}}} .dy + \dfrac{1}{2}\int {\dfrac{{2y}}{{1 + {y^2}}}} .dy + \int {\dfrac{1}{{1 + {y^2}}}} .dy \\
   = - \log \left| {1 - y} \right| + \dfrac{1}{2}\log \left| {1 + {y^2}} \right| + {\tan ^{ - 1}}y + C \\
   = \log \left| {\dfrac{{\sqrt {1 + {y^2}} }}{{1 - y}}} \right| + {\tan ^{ - 1}}y + C \\
   = \log \dfrac{{\sqrt {1 + {{\sin }^2}x} }}{{1 - \sin x}} + {\tan ^{ - 1}}\left( {\sin x} \right) + C \;
  $
Therefore, the required answer is $ \log \dfrac{{\sqrt {1 + {{\sin }^2}x} }}{{1 - \sin x}} + {\tan ^{ - 1}}\left( {\sin x} \right) + C $
So, the correct answer is “$ \log \dfrac{{\sqrt {1 + {{\sin }^2}x} }}{{1 - \sin x}} + {\tan ^{ - 1}}\left( {\sin x} \right) + C $”.

Note: There are different ways to solve a limit. All different ways have different answers. But all those answers are correct as long as the method and the steps are done correctly without any mistakes. Here in this question, we used the method of partial fraction to solve the limit. The rational equation and its partial fraction is an identity which means that it is a statement which is true for all permissible values of x. Fun fact- the equal sign is used to indicate that the statement is an equation meaning to indicate that the statement is true only for certain values of x.