Question

Choose the correct option: \[\int_0^{\dfrac{\pi }{4}} {\frac{{\sec x}}{{(1 + \tan x)(2 + \tan x)}}} \]
A. ${\log _e}\left( {\dfrac{2}{3}} \right)$
B. ${\log _e}3$
C. $\dfrac{1}{2}{\log _e}\left( {\dfrac{4}{3}} \right)$
D. ${\log _e}\left( {\dfrac{4}{3}} \right)$

Answer 1

Hint: Here given an integral of a function of which we have to find the integration of the function, here the function is called the integrand. There are two types of integrals, which are definite integrals and indefinite integrals. Definite integrals are those integrals for which there is a particular limit on the integrals. Whereas for the indefinite integrals there are no such limits on the integral. This integration is done by the method of substitution and partial fractions.

Step-By-Step answer:
There are 3 different methods of solving integrations which are:
> Integration by substitution.
> Integration by partial fractions.
> Integration by parts.
Here we are using the first two methods of integrations in order to solve the given integral.
\[\int_0^{\dfrac{\pi }{4}} {\dfrac{{\sec x}}{{(1 + \tan x)(2 + \tan x)}}} \]
Now applying the method of integration by substitution.
Let $\tan x = t$
Now differentiate the above expression on both sides:
$ \Rightarrow {\sec ^2}xdx = dt$
Hence the upper and the lower limits become:
$\tan x = t$
As $x = 0$$ \Rightarrow \tan 0 = 0$
$\therefore $Lower limit : $t = 0$
As \[x = \dfrac{\pi }{4} \Rightarrow \tan \dfrac{\pi }{4} = 1\]
$\therefore $Upper limit : $t = 1$
Substituting the above two expressions in the given definite integral gives:
$ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt$
Now applying the method of integration by partial fractions.
Consider \[\dfrac{1}{{(1 + t)(2 + t)}}\], now splitting this complex fraction into partial fractions:
$ \Rightarrow \dfrac{1}{{(1 + t)(2 + t)}} = \dfrac{A}{{1 + t}} + \dfrac{B}{{2 + t}}$
Simplifying the R.H.S:
$ \Rightarrow \dfrac{1}{{(1 + t)(2 + t)}} = \dfrac{{A(2 + t) + B(1 + t)}}{{(1 + t)(2 + t)}}$
$ \Rightarrow \dfrac{1}{{(1 + t)(2 + t)}} = \dfrac{{(A + B)t + 2A + B}}{{(1 + t)(2 + t)}}$
On comparing the both sides, we get the values of A and B:
$ \Rightarrow A + B = 0$
$ \Rightarrow 2A + B = 1$
We have two equations and two variables, hence obtaining the values of A and B:
$ \Rightarrow A = 1;B = - 1$
Substituting these values in the integral, as given by:
$ \Rightarrow \dfrac{1}{{(1 + t)(2 + t)}} = \dfrac{1}{{1 + t}} + \dfrac{{ - 1}}{{2 + t}}$
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = \int\limits_0^1 {\left( {\dfrac{1}{{1 + t}} + \dfrac{{ - 1}}{{2 + t}}} \right)} dt\]
Splitting the integrals on the R.H.S:
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = \int\limits_0^1 {\dfrac{1}{{1 + t}}} dt + \int\limits_0^1 {\dfrac{{ - 1}}{{2 + t}}} dt\]
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = {\log _e}(1 + t)_0^1 - {\log _e}(2 + t)_0^1 + c\]
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = {\log _e}\left( {\dfrac{{1 + t}}{{2 + t}}} \right)_0^1 + c\]
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = {\log _e}\left( {\dfrac{{1 + 1}}{{2 + 1}}} \right) - {\log _e}\left( {\dfrac{{1 + 0}}{{2 + 0}}} \right) + c\]
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = {\log _e}\left( {\dfrac{2}{3}} \right) - {\log _e}\left( {\dfrac{1}{2}} \right) + c\]
Applying the basic formula from logarithms ${\log _e}a - {\log _e}b = {\log _e}\left( {\dfrac{a}{b}} \right)$
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = {\log _e}\dfrac{{\left( {\dfrac{2}{3}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} + c\]
\[ \Rightarrow \int\limits_0^1 {\dfrac{1}{{(1 + t)(2 + t)}}} dt = {\log _e}\left( {\dfrac{4}{3}} \right) + c\]

\[\int_0^{\dfrac{\pi }{4}} {\dfrac{{\sec x}}{{(1 + \tan x)(2 + \tan x)}}} = {\log _e}\left( {\dfrac{4}{3}} \right) + c\]

Note: Please note that after applying the integration by partial fractions and solving the integral we can substitute back the value of $t = \tan x$, and the limits of x, will give the same final answer. It can be done in either of the ways, but finally we end up getting the same solution.