Question

The value of $\int_0^\pi {\log \left( {1 + \cos x} \right)dx} $ is
A. $ - \dfrac{\pi }{2}\log 2$
B. $\pi \log \dfrac{1}{2}$
C. $ - \pi \log 2$
D. $\dfrac{\pi }{2}\log 2$

Answer 1

Hint: We have to solve this question step by step by assuming the integration values as variables. By simple integration method by using the properties of integration and logarithms. Then, we also use a bit of derivation and eventually substitute all the real values in the assumed variables to get the answer by putting the limits in our calculated result.

Complete step-by-step answer:
Let us assume $I = \int_0^\pi {\log \left( {1 + \cos x} \right)dx} $…… (1)
Now, using identity $\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} $ , it can be rewritten as
$I = \int_0^\pi {\log \left( {1 + \cos \left( {\pi - x} \right)} \right)dx} $
Now, as we know $\cos \left( {\pi - x} \right) = - \cos x$
$I = \int_0^\pi {\log \left( {1 - \cos x} \right)dx} $…… (2)
Now, adding (1) and (2) we get
\[
  2I = \int_0^\pi {\log \left( {1 + \cos x} \right)dx} + \int_0^\pi {\log \left( {1 - \cos x} \right)dx} \\
  2I = \int_0^\pi {\log \left( {1 + \cos x} \right)dx} + \log \left( {1 - \cos x} \right)dx \\
 \]
Here, we use the multiplication property of log i.e. log(a)+log(b) = log(a.b)
$
  2I = \int_0^\pi {\log \left[ {\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)} \right]dx} \\
  2I = \int_0^\pi {\log \left[ {\left( {1 - {{\cos }^2}x} \right)} \right]dx} \\
 $
Now, we know that $1 - {\cos ^2}x = {\sin ^2}x$
So, $2I = \int_0^\pi {\log \left( {{{\sin }^2}x} \right)dx} $
Now, we use another property of log i.e. $\log {a^b} = b\log a$
$
  2I = \int_0^\pi {2\log \left( {\sin x} \right)dx} \\
  2I = 2\int_0^\pi {\log \left( {\sin x} \right)dx} \\
  I = \int_0^\pi {\log \left( {\sin x} \right)dx} \\
 $
To solve this, we need to use another property of integration which is
$\int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx} } $ if $f\left( {2a - x} \right) = f\left( x \right)$
Here, $f\left( x \right) = \log \sin x$
$
  f\left( {\pi - x} \right) = \log \left[ {\sin \left( {\pi - x} \right)} \right]dx = \log \left( {\sin x} \right)dx = f\left( x \right) \\
  \therefore I = \int_0^\pi {\log \left( {\sin x} \right)dx = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)dx} } \\
 $
Now, again assuming $J = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)dx} $ ……(3)
For solving $J$ we use the property $ - \int_0^a {f\left( x \right)dx = \int_0^a {f\left( {a - x} \right)dx} } $
$\therefore J = \int_0^{\dfrac{\pi }{2}} {\log \left[ {\sin \left( {\dfrac{\pi }{2} - x} \right)} \right]dx} $
$J = \int_0^{\dfrac{\pi }{2}} {\log \left( {\cos x} \right)dx} $ …. (4)
Adding (3) and (4)
$2J = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)dx} + \int_0^{\dfrac{\pi }{2}} {\log \left( {\cos x} \right)dx} $
Again by using the multiplication property of log
$2J = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x.\cos x} \right)dx} $
Now, divide and multiply by 2
$2J = \int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{2\sin x.\cos x}}{2}} \right)dx} $
Now, using the division property of log i.e. $\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)$
$2J = \int_0^{\dfrac{\pi }{2}} {\left[ {\log \left( {\sin 2x} \right) - \log 2} \right]dx} $
$2J = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin 2x} \right)dx - \int_0^{\dfrac{\pi }{2}} {\log \left( 2 \right)dx} } $ …. (5)
Now, assuming $K = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin 2x} \right)dx} $
Let 2x =t
Differentiating on both sides
$
  2 = \dfrac{{dt}}{{dx}} \\
  dx = \dfrac{{dt}}{2} \\
 $
Now, finding the limits
If x = 0, then value of t = 2(0) = 0
If x = $\dfrac{\pi }{2}$, then value of t = 2($\dfrac{\pi }{2}$) = $\pi $
$\therefore $ Putting the values of t and $dt$ and changing the limits,
$
  K = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin 2x} \right)dx} \\
  K = \int_0^\pi {\log \left( {\sin t} \right)\dfrac{{dt}}{2}} \\
  K = \dfrac{1}{2}\int_0^\pi {\log \left( {\sin t} \right)dt} \\
 $
Again using this property,
$\int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx} } $ if $f\left( {2a - x} \right) = f\left( x \right)$
$
  f\left( t \right) = \log \sin t \\
  f(2a - t) = f\left( {2\pi - t} \right) = \log \sin t \\
 $
$\therefore K = \dfrac{1}{2}\int_0^\pi {\log \sin tdt = \dfrac{1}{2} \times 2\int_0^{\dfrac{\pi }{2}} {\log \sin tdt = } } \int_0^{\dfrac{\pi }{2}} {\log \sin tdt} $
Now, using the property
$\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( t \right)dt} $
 \[K = \int_0^{\dfrac{\pi }{2}} {\log \sin x dx} \]
Substituting value of $K$ in (5)
$2J = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x} \right)dx - \int_0^{\dfrac{\pi }{2}} {\log \left( 2 \right)dx} } $
Now, by using (3) we get
$
  2J = J - \log \left( 2 \right)\int_0^{\dfrac{\pi }{2}} {1.dx} \\
  2J - J = - \log \left( 2 \right)\left[ x \right]_0^{\dfrac{\pi }{2}} \\
  J = - \log 2\left[ {\dfrac{\pi }{2} - 0} \right] \\
  J = - \log 2\left[ {\dfrac{\pi }{2}} \right] \\
  J = - \dfrac{\pi }{2}\log 2 \\
 $
Hence, $I = 2J = 2 \times - \dfrac{\pi }{2}\log 2 = - \pi \log 2$
Correct option is C.

Note: This is a very important question if you want to learn integration. Solving integration by using its properties is done in the question three times. Always remember the properties of integration and solve the question very carefully as every step can be very confusing. Thus, a simple mistake will lead to a lot of ambiguities. Therefore, this is the only way to reach for the answer.