Question

The following integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {{{\left( {2\cos ec{\text{ }}x} \right)}^{17}}} dx$ is equal to
A) \[\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}du} \]
B) \[\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} + {e^{ - u}}} \right)}^{17}}du} \]
C) \[\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} - {e^{ - u}}} \right)}^{17}}du} \]
D) \[\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} - {e^{ - u}}} \right)}^{16}}du} \]

Answer 1

Hint: In the integral calculus, we are to find a function whose differential is given. Thus, integration is a process that is the inverse of differentiation.
Let $\dfrac{{dF\left( x \right)}}{{dx}} = f\left( x \right)$, then $\smallint f\left( x \right)dx = F\left( x \right) + C$ , where C is the integration constant.
The given question is not about solving the integral but to change the form of the integral.
The Options of the given question includes the integral in the form of exponential $\left( e \right)$ . Therefore try to substitute the given integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {{{\left( {2\cos ec{\text{ }}x} \right)}^{17}}} dx$ in exponential as well and change the limits accordingly.

Complete step-by-step answer:
Step 1: Find the Euler form of the integral
The given integral integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {{{\left( {2\cos ec{\text{ }}x} \right)}^{17}}} dx$
Let \[\cos ec{\text{ }}x - \cot x = {e^t}\] …… (1)
Trigonometric Identity: \[\cos e{c^2}x - {\cot ^2}x = 1\]
Expand using the identity: ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Take \[\cos ec{\text{ }}x = a,{\text{ }}\cot x = b\]
Therefore, \[\cos e{c^2}x - {\cot ^2}x = \] \[\left( {\cos ec{\text{ }}x - \cot x} \right)\left( {\cos ec{\text{ }}x + \cot x} \right) = 1\]
But \[\left( {\cos ec{\text{ }}x - \cot x} \right) = {e^t}\] , substitute it from equation (1).
\[
   \Rightarrow \left( {{e^t}} \right)\left( {\cos ec{\text{ }}x + \cot x} \right) = 1 \\
   \Rightarrow \left( {\cos ec{\text{ }}x + \cot x} \right) = \dfrac{1}{{{e^t}}} \\
 \]
\[ \Rightarrow \cos ec{\text{ }}x + \cot x = {e^{ - t}}\] …… (2)
Adding equations (1) and (2)
\[\left( {\cos ec{\text{ }}x - \cot x + \cos ec{\text{ }}x + \cot x} \right) = {e^t} + {e^{ - t}}\]
\[ \Rightarrow 2\cos ec{\text{ }}x = {e^t} + {e^{ - t}}\]
Or \[\cos ec{\text{ }}x = \dfrac{{{e^t} + {e^{ - t}}}}{2}\] …… (3)
Differentiating both sides of equation (1)
\[\cos ec{\text{ }}x - \cot x = {e^t}\]
Using the differentiation formulas for: $\dfrac{{d\left( {\cos ec{\text{ x}}} \right)}}{{dx}} = - \left( {\cos ec{\text{ }}x} \right)\left( {\cot x} \right)$ ,
$\dfrac{{d\left( {\cot x} \right)}}{{dx}} = - \cos e{c^2}x$ ,
$\dfrac{{\operatorname{d} \left( {{e^t}} \right)}}{{dt}} = {e^t}$
\[ \Rightarrow \left( { - \left( {\cos ec{\text{ }}x} \right) \times \cot x + \cos e{c^2}x} \right)dx = {e^t}dt\]
Taking $\cos ec{\text{ x}}$as common
\[ \Rightarrow \cos ec{\text{ }}x\left( {\cos ec{\text{ }}x - \cot x} \right)dx = {e^t}dt\]
Substituting \[\cos ec{\text{ }}x\] and \[\left( {\cos ec{\text{ }}x - \cot x} \right)\] from equation (3) and (1) respectively.
\[ \Rightarrow \left( {\dfrac{{{e^t} + {e^{ - t}}}}{2}} \right){e^t}dx = {e^t}dt\]
\[\because dx = \dfrac{{2dt}}{{{e^t} + {e^{ - t}}}}\] …... (4)
So far we have changed the integral $\cos ec{\text{ }}x$ (i.e. equation 3) and the differential $dx$ (i.e. equation 4) in terms of exponential as given the options.
Step 2: Change of limits
For $x = \dfrac{\pi }{4}$
$
  \therefore {e^t} = \cos ec{\text{ }}\left( {\dfrac{\pi }{4}} \right) - \cot \left( {\dfrac{\pi }{4}} \right) \\
   \Rightarrow \sqrt 2 - 1 \\
  \therefore t = \log \left( {\sqrt 2 - 1} \right) \\
 $
But in the options, limits are given in the form of $\log \left( {\sqrt 2 + 1} \right)$
Change the values $\log \left( {\sqrt 2 - 1} \right)$ using the property of logarithmic:
$a\log b = \log {b^a}$
We can write $t = \left( { - 1} \right)\left( { - 1} \right)\log \left( {\sqrt 2 - 1} \right)$
$
   \Rightarrow \left( { - 1} \right)\log {\left( {\sqrt 2 - 1} \right)^{ - 1}} \\
   \Rightarrow \left( { - 1} \right)\log \left( {\dfrac{1}{{\sqrt 2 - 1}}} \right) \\
 $
On rationalizing $\dfrac{1}{{\sqrt 2 - 1}}$
$
   \Rightarrow \dfrac{1}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}} \\
   \Rightarrow \dfrac{{\sqrt 2 + 1}}{{2 - 1}} = \sqrt 2 + 1 \\
 $
Hence, $t = \left( { - 1} \right)\log \left( {\sqrt 2 + 1} \right)$

For $x = \dfrac{\pi }{2}$
$
  \therefore {e^t} = \cos ec{\text{ }}\left( {\dfrac{\pi }{2}} \right) - \cot \left( {\dfrac{\pi }{2}} \right) \\
   \Rightarrow 1 - 0 = 1 \\
  \therefore t = \ln \left( 1 \right) = 0 \\
 $
Therefore, the given integral becomes:
On changing limits and substituting equations (3) and (4)
\[\int\limits_{ - \ln \left( {1 + \sqrt 2 } \right)}^0 {{{\left( {{e^t} + {e^{ - t}}} \right)}^{17}}\dfrac{{2dt}}{{{e^t} + {e^{ - t}}}}} \]
\[ \Rightarrow \int\limits_{ - \ln \left( {1 + \sqrt 2 } \right)}^0 {2{{\left( {{e^t} + {e^{ - t}}} \right)}^{16}}dt} \]
Step 3: Change the limits according to the options.
Take $t = - u$
Differentiating both sides
$dt = - du$
Change of limits:
$
  t = - \log \left( {\sqrt 2 + 1} \right) \Rightarrow u = - t = \log \left( {\sqrt 2 + 1} \right) \\
  t = 0 \Rightarrow u = 0 \\
 $
On substituting the integral becomes:
\[ - \int\limits_{\ln \left( {1 + \sqrt 2 } \right)}^0 {2{{\left( {{e^{ - u}} + {e^{ - \left( { - u} \right)}}} \right)}^{16}}du} \]
Using the property of definite integral
$\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx$ i.e. when upper limits and lower limits are interchanged, the integral is multiplied by a minus sign, .
\[ \Rightarrow \int\limits_0^{\ln \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^{ - u}} + {e^u}} \right)}^{16}}du} \]
Or \[\int\limits_0^{\ln \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}du} \]
On simplification of the given integral, it comes out to be\[\int\limits_0^{\ln \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}du} \] .

Thus, the correct option is (A).

Note: Carefully do the calculation, emphasize on each and every step especially while changing
limits of the integral.
Similarly, if the given question was integral to $\sec x$.
You should substitute $\sec x - \tan x = {e^t}$. With the use of trigonometric identity ${\sec ^2}x - {\tan ^2}x = 1$ , you can easily calculate $\sec x - \tan x$ similarly as calculated above.