Question

If an= \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 - \cos 2nx}}{{1 - \cos 2x}}} dx\],then a1, a2, a3, …. an are in
1)A.P
2)G.P
3)H.P
4)All of these

Answer 1

Hint: Since the above question cannot be integrated directly by using integration formulas, we need to use trigonometric identities and formulas to simplify the question and bring it in the form where we can integrate easily using integration formulas and properties. Try to solve the integration for values of n starting from 1.

Complete step-by-step answer:
We have to find the value of the given integral an = \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 - \cos 2nx}}{{1 - \cos 2x}}} dx\]
For a1, a2, a3, …. an
We will start solving the problem by using the trigonometric identity $ 1 - \cos 2x = 2{\sin ^2}x $
Hence our given integral will be
an = \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 - \cos 2nx}}{{1 - \cos 2x}}} dx\]= \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{2{{\sin }^2}nx}}{{2{{\sin }^2}x}}} dx = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}nx}}{{{{\sin }^2}x}}} dx\]
hence, \[{{\text{a}}_n} = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}nx}}{{{{\sin }^2}x}}} dx\] respectively.
Now we will start solving the integral for every value of n
For n=1, we get
 $
  {a_1} = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x}}dx} \\
   = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \\
   = \left[ x \right]_0^{\dfrac{\pi }{2}} \\
   = \dfrac{\pi }{2} \;
  $
hence $ {a_1} = \dfrac{\pi }{2} $
for n=2 we get
 $ {a_2} = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}2x}}{{{{\sin }^2}x}}dx} $
 $ = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin 2x.\sin 2x}}{{{{\sin }^2}x}}dx} $
  $ = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {2\sin x\cos x} \right).\left( {2\sin x\cos x} \right)}}{{{{\sin }^2}x}}dx} - - - $ using $ \sin 2\theta = 2\sin \theta \cos \theta $
 $ = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{4{{\sin }^2}x.{{\cos }^2}x}}{{{{\sin }^2}x}}} $
Further solving we get
 $ {a_2} = 4\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} $
We know that $ 2{\cos ^2}x = \cos 2x + 1 $
Hence we get
 $
  {a_2} = 4\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos 2x + 1}}{2}dx} \\
   = 2\int\limits_0^{\dfrac{\pi }{2}} {\cos 2x + 1dx} \\
  $
Further integrating we get
\[
  {a_2} = 2\left[ {\left[ {\dfrac{{\sin 2x}}{2}} \right]_0^{\dfrac{\pi }{2}} + \left[ x \right]_0^{\dfrac{\pi }{2}}} \right] \\
   = \sin \pi + \pi \\
   = \pi \;
 \]
Hence \[{a_2} = \pi \]
Now for n=3
 $ {a_3} = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}3x}}{{{{\sin }^2}x}}dx} $
Solving this integral we get
$ {a_3} = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\left( {3\sin x - 4{{\sin }^3}x} \right)}^2}}}{{{{\sin }^2}x}}dx} $ since $ \sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta $
 $
{a_3} = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\left( {3\sin x - 4{{\sin }^3}x} \right)}^2}}}{{{{\sin }^2}x}}dx} \\
   = \int\limits_0^{\dfrac{\pi }{2}} {{{\left( {\dfrac{{3\sin x - 4{{\sin }^3}x}}{{\sin x}}} \right)}^2}dx} \\
   = \int\limits_0^{\dfrac{\pi }{2}} {{{\left( {3 - 4{{\sin }^2}x} \right)}^2}dx} \;
  $
Using $ 1 - \cos 2x = 2{\sin ^2}x $ we get
 $
  {a_3} = \int\limits_0^{\dfrac{\pi }{2}} {{{\left( {3 - 2(1 - \cos 2x)} \right)}^2}dx} \\
   = \int\limits_0^{\dfrac{\pi }{2}} {{{\left( {3 - 2 + 2\cos 2x} \right)}^2}dx} \\
   = \int\limits_0^{\dfrac{\pi }{2}} {{{\left( {1 + 2\cos 2x} \right)}^2}dx} \;
  $
further using solving we get
 $
  {a_3} = \int\limits_0^{\dfrac{\pi }{2}} {1 + 4{{\cos }^2}2x + 4\cos 2xdx} \\
  = \left[ x \right]_0^{\dfrac{\pi }{2}} + 4\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos 4x + 1}}{2}} + 4\dfrac{{\left[ {\sin 2x} \right]_0^{\dfrac{\pi }{2}}}}{2} \\
  $ since $ 2{\cos ^2}x = \cos 2x + 1 $
 $
 = \dfrac{\pi }{2} + 2\left[ {\dfrac{{\sin 4x}}{4}} \right]_0^{\dfrac{\pi }{2}} + 2\left[ x \right]_0^{\dfrac{\pi }{2}} + 2\sin 2\pi \\
   = \dfrac{\pi }{2} + \dfrac{1}{2}\sin 2\pi + \pi + 0 \\
   = \dfrac{\pi }{2} + \pi \\
   = \dfrac{{3\pi }}{2} \;
  $
Hence $ {a_3} = \dfrac{{3\pi }}{2} $
Therefore , $ {a_1} = \dfrac{\pi }{2},{a_2} = \pi ,{a_3} = \dfrac{{3\pi }}{2} $
Since the difference between the consecutive term in the above sequence is constant, hence the given term is in AP
Hence the right option to above question is option (A)

Note: Whenever asked for the sequence to be found in AP, GP, or HP, try to solve the integral for minimum three terms. When given trigonometric forms which cannot be integrated directly, use trigonometric identities which are related and help the problem to get simplified and not vice versa.