Question

If $ \int {\dfrac{{\tan x}}{{1 + \tan x + {{\tan }^2}x}}} dx = x - \dfrac{K}{{\sqrt A }}{\tan ^{ - 1}}\left( {\dfrac{{K\tan x + 1}}{{\sqrt A }}} \right) + C $ , (C is a constant of integration). Find the ordered pair $ \left( {K,A} \right) $
(A) $ \left( {2,3} \right) $
(B) $ \left( {2,1} \right) $
(C) $ \left( { - 2,1} \right) $
(D) $ \left( { - 2,3} \right) $

Answer 1

Hint: Take the integral from the left side of the given equation and solve it. Add and subtract $ \left( {1 + {{\tan }^2}x} \right) $ from the numerator and then separate the integral into two parts. Split the numerator as: $ \left( {\tan x + 1 + {{\tan }^2}x} \right) - \left( {1 + {{\tan }^2}x} \right) $ . By this way, the first integral will become $ 1dx $ and for the second integral use substitution $ \tan x = m $ and solve it further. After solving, compare the coefficients with the right side of the given equation.

Complete step-by-step solution:
Here we need to solve the integral on the left side of the given equation and then compare the solution with the right side of the equation. And find the values of the coefficients in the right side.
Let’s take the right side and try to evaluate the integral.
By adding and subtracting the term $ \left( {1 + {{\tan }^2}x} \right) $ in the numerator, we get:
 $ \Rightarrow \int {\dfrac{{\tan x}}{{1 + \tan x + {{\tan }^2}x}}} dx = \int {\dfrac{{\tan x + 1 + {{\tan }^2}x - \left( {1 + {{\tan }^2}x} \right)}}{{1 + \tan x + {{\tan }^2}x}}} dx $
Now this will let us separate this one integral into two parts and solve them separately
 $ \Rightarrow \int {\dfrac{{\tan x + 1 + {{\tan }^2}x - \left( {1 + {{\tan }^2}x} \right)}}{{1 + \tan x + {{\tan }^2}x}}} dx = \int {\dfrac{{\tan x + 1 + {{\tan }^2}x}}{{1 + \tan x + {{\tan }^2}x}}} dx - \int {\dfrac{{\left( {1 + {{\tan }^2}x} \right)}}{{1 + \tan x + {{\tan }^2}x}}} dx $
We can see that the first integral becomes $ \int {1dx} $ and for the second one we can use the trigonometric identity $ 1 + {\tan ^2}x = {\sec ^2}x $ in the numerator. This will give us:
 $ \Rightarrow \int {\dfrac{{\tan x + 1 + {{\tan }^2}x}}{{1 + \tan x + {{\tan }^2}x}}} dx - \int {\dfrac{{\left( {1 + {{\tan }^2}x} \right)}}{{1 + \tan x + {{\tan }^2}x}}} dx = \int {1dx - \int {\dfrac{{{{\sec }^2}x}}{{1 + \tan x + {{\tan }^2}x}}} } dx $
Let’s use the substitution method to solve the second integral. Assume $ \tan x = m \Rightarrow {\sec ^2}xdx = dm $
 $ \Rightarrow \int {1dx - \int {\dfrac{{{{\sec }^2}x}}{{1 + \tan x + {{\tan }^2}x}}} } dx = x - \int {\dfrac{{dm}}{{1 + m + {m^2}}}} $
Now, using the identity $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ in the denominator for completing the square in the quadratic function. For this, we need to add and subtract $ \dfrac{1}{4} $ in the denominator and also multiply and divide with $ 2 $ in the term $ 'm' $
 $ \Rightarrow 1 + m + {m^2} = 1 + 2 \times \dfrac{1}{2} \times m + {m^2} + \dfrac{1}{4} - \dfrac{1}{4} = {\left( {m + \dfrac{1}{2}} \right)^2} + 1 - \dfrac{1}{4} = {\left( {m + \dfrac{1}{2}} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} $
Therefore, we can now put $ 1 + m + {m^2} = {\left( {m + \dfrac{1}{2}} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} $ in the denominator
 $ \Rightarrow x - \int {\dfrac{{dm}}{{1 + m + {m^2}}}} = x - \int {\dfrac{{dm}}{{{{\left( {m + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}} $
As we know the integration formula $ \int {\dfrac{{dx}}{{{a^2} + {x^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C $ , which can be used here with $ x = m + \dfrac{1}{2}{\text{ and }}a = \dfrac{{\sqrt 3 }}{2} $
 $ \Rightarrow x - \int {\dfrac{{dm}}{{{{\left( {m + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}} = x - \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{{\left( {m + \dfrac{1}{2}} \right)}}{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}} = x - \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{2m + 1}}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right) = x - \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2m + 1}}{{\sqrt 3 }}} \right) + C $ Now, we can put back our substitution $ \tan x = m $ into this solution.
Therefore, we get: $ \int {\dfrac{{\tan x}}{{1 + \tan x + {{\tan }^2}x}}} dx = x - \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2\tan x + 1}}{{\sqrt 3 }}} \right) + C $
We can now compare the coefficient of our solution for this integral with the given equation, i.e. $ \int {\dfrac{{\tan x}}{{1 + \tan x + {{\tan }^2}x}}} dx = x - \dfrac{K}{{\sqrt A }}{\tan ^{ - 1}}\left( {\dfrac{{K\tan x + 1}}{{\sqrt A }}} \right) + C $
Thus, we can clearly see that $ K = 2{\text{ and }}A = 3 $ and the required ordered pair $ \left( {K,A} \right) = \left( {2,3} \right) $

Hence, the option (A) is the correct answer

Note: In questions like this, transforming the solution of the integration into the required form to compare with the given expression is also an important part. Do not forget to add a constant ‘C’ after the integration to compensate for a zero term. Since we know that the differentiation of any constant is zero. Therefore, a solution of indefinite integration should have a constant term in it.