Question

Evaluate the value of $\int_0^\pi {\log \left( {1 + \cos x} \right)} dx$

Answer 1

Hint: - Use the property of definite integral $\int_0^{2a} {f\left( x \right)dx = 2} \int_0^a {f\left( x \right)dx} ,{\text{ if }}f\left( {2a - x} \right) = f\left( x \right)$
, and $\int_0^a {f\left( t \right)} dt = \int_0^a {f\left( {a - t} \right)} dt$. Definite integral is the one which has upper and lower limits whereas indefinite integral no upper and lower limits are there.

Let $I = \int_0^\pi {\log \left( {1 + \cos x} \right)} dx$
As we know $\left( {1 + \cos x} \right) = 2{\cos ^2}\left( {\dfrac{x}{2}} \right)$
Substitute this value in the integral
${\text{I = }}\int_0^\pi {\log \left( {2{{\cos }^2}\left( {\dfrac{x}{2}} \right)} \right)dx}$
As we know,$\log \left( {ab} \right) = \log a + \log b$, so apply this property

$$\Rightarrow {\text{I = }}\int_0^\pi {\left( {\log \left( {{{\cos }^2}\left( {\dfrac{x}{2}} \right)} \right) + \log 2} \right)dx} \\\ \Rightarrow I = \int_0^\pi {\log \left( {{{\cos }^2}\left( {\dfrac{x}{2}} \right)} \right)} dx + \int_0^\pi {\log 2} dx \\\$$

Now we know $\log {a^2} = 2\log a$so apply this property

\int_0^\pi {\log 2} dx$$ Now, let $$\dfrac{x}{2} = t.................\left( 1 \right)$$ If $$x = 0 \Rightarrow t = 0$$ If $$x = \pi \Rightarrow t = \dfrac{\pi }{2}$$ Differentiate equation (1) w.r.t.$x$ $$ \Rightarrow dx = 2dt$$ Substitute these values in the integral

\Rightarrow I = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos \left( t \right)} \right)} 2dt +
\int_0^\pi {\log 2} dx \\
\Rightarrow I = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos \left( t \right)} \right)} dt +
\int_0^\pi {\log 2} dx \\
\Rightarrow I = {I_1} + {I_2} \\

Now first solve $${I_1}$$ $$ \Rightarrow {I_1} = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos \left( t \right)} \right)} dt.................\left( 2 \right)$$ As we know $$\int_0^a {f\left( t \right)} dt = \int_0^a {f\left( {a - t} \right)} dt............\left( 3 \right)$$ Apply this definite integral property in $${I_1}$$

\Rightarrow {I_1} = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos \left( {\dfrac{\pi }{2} - t}
\right)} \right)} dt \\

\Rightarrow {I_1} = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin t} \right)} dt.....................\left(
4 \right) \\

$$Now add equation (2) and (4)$$

\Rightarrow 2{I_1} = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos t} \right)} dt +
4\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin t} \right)} dt \\
\Rightarrow 2{I_1} = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos t \times \sin t} \right)} dt
\\

As we know $$2\cos t \times \sin t = \sin 2t$$so apply this

\Rightarrow 2{I_1} = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}\sin 2t} \right)} dt
\\
\Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}\sin 2t} \right)} dt
\\
\Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt +
2\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin 2t} \right)} dt \\

Let, $${\text{2t = v}}...........\left( 5 \right)$$ If, $${\text{ t = 0}} \Rightarrow {\text{v = 0}}$$ If, $${\text{t = }}\dfrac{\pi }{2} \Rightarrow v = \pi $$ Now, differentiate equation (5) w.r.t.$t$ $$ \Rightarrow {\text{2dt = dv}}$$ So substitute these values in the integral

\Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt +
2\int_0^\pi {\log \left( {\sin v} \right)} \dfrac{{dv}}{2} \\
\Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt + \int_0^\pi
{\log \left( {\sin v} \right)} dv \\

As we know $$\int_0^{2a} {f\left( x \right)dx = 2} \int_0^a {f\left( x \right)dx} ,{\text{ if }}f\left( {2a - x} \right) = f\left( x \right)$$ So, comparing from above equation $2a = \pi , \Rightarrow a = \dfrac{\pi }{2}$ $$ \Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt + 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin v} \right)} dv$$ Now from equation (3)

\Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt +
2\int_0^{\dfrac{\pi }{2}} {\log \left( {\sin \left( {\dfrac{\pi }{2} - v} \right)} \right)} dv \\
\Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt +
2\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos v} \right)} dv \\

As we know in definite integral we change the variable so we change the variable to $t$ in the second integral in the above equation. $$ \Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt + 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\cos t} \right)} dt$$ From equation (2)

\Rightarrow {I_1} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt +
\dfrac{{{I_1}}}{2} \\
\Rightarrow \dfrac{{{I_1}}}{2} = 2\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt
\\
\Rightarrow {I_1} = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt \\
I = {I_1} + {I_2} \\
\Rightarrow I = 4\int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{1}{2}} \right)} dt + \int_0^\pi
{\log 2dx} \\
\Rightarrow I = 4\log \left( {\dfrac{1}{2}} \right)\left[ t \right]_0^{\dfrac{\pi }{2}} + \log
2\left[ x \right]_0^\pi \\
\Rightarrow I = 4\log \left( {\dfrac{1}{2}} \right)\left[ {\dfrac{\pi }{2}} \right] + \log 2\left[
\pi \right] \\
\Rightarrow I = 2\pi \log \left( {\dfrac{1}{2}} \right) + \pi \log 2 \\

Now, as we know $$\log \left( {\dfrac{1}{2}} \right) = - \log 2$$ $$ \Rightarrow I = - 2\pi \log 2 + \pi \log 2{\text{ }} \Rightarrow I = - \pi \log 2$$ So, this is the required value of the integral. Note: - In such types of questions the key concept we have to remember is that always remember all the properties of definite integral which is stated above, then using this properties simplify the integral and use some base logarithmic and trigonometry properties which is also stated above, then simplify we will get the required answer.