Answer

Verified

489k+ views

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

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

\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.

\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

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

\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.

Recently Updated Pages

Who among the following was the religious guru of class 7 social science CBSE

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Trending doubts

Derive an expression for drift velocity of free electrons class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE

The energy of a charged conductor is given by the expression class 12 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Derive an expression for electric field intensity due class 12 physics CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Derive an expression for electric potential at point class 12 physics CBSE