Question

integration 0 - pie/2 log (cos x) dx ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_0^{\dfrac{\pi}{2} }\sf log(cosx) \: dx \:$

Let assume that

$\rm :\longmapsto\:I_1 \: = \: \displaystyle\int_0^{\dfrac{\pi}{2} }\sf log(cosx) \: dx \: - - - (1)$

We know,

$\boxed{ \bf{ \: \displaystyle\int_0^{a}\sf \: f(x) \: dx \: = \: \displaystyle\int_0^{a}\sf f(a - x) \: dx}}$

So, using this property, we get

$\rm :\longmapsto\:I_1 \: = \: \displaystyle\int_0^{\dfrac{\pi}{2} }\sf logcos \bigg(\dfrac{\pi}{2} - x \bigg) \: dx \:$

$\rm :\longmapsto\:I_1 \: = \: \displaystyle\int_0^{\dfrac{\pi}{2} }\sf log(sinx) \: dx \: - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2I_1 \: = \:\displaystyle\int_0^{\dfrac{\pi}{2}}\sf log(cosx)dx + \displaystyle\int_0^{\dfrac{\pi}{2} }\sf log(sinx) \: dx \:$

$\rm :\longmapsto\:2I_1 \: = \:\displaystyle\int_0^{\dfrac{\pi}{2}}\sf \bigg( logcosx + logsinx \bigg) \: dx$

$\rm :\longmapsto\:2I_1 \: = \:\displaystyle\int_0^{\dfrac{\pi}{2}}\sf \bigg( log(cosx sinx) \bigg) \: dx$

$\red{\bigg \{ \because \:logx + logy = logxy \bigg \}}$

$\rm :\longmapsto\:2I_1 = \displaystyle\int_0^{\dfrac{\pi}{2}}\sf log \bigg(\dfrac{2sinxcosx}{2} \bigg)$

$\rm :\longmapsto\:2I_1 = \displaystyle\int_0^{\dfrac{\pi}{2}}\sf log \bigg(\dfrac{sin2x}{2} \bigg)$

$\rm :\longmapsto\:2I_1 = \displaystyle\int_0^{\dfrac{\pi}{2}}\sf (log sin2x - log2)\: dx \:$

$\red{\bigg \{ \because \: log\dfrac{x}{y} = logx - logy\bigg \}}$

$\rm :\longmapsto\:2I_1 = \displaystyle\int_0^{\dfrac{\pi}{2}}\sf log sin2x \: dx - log2\displaystyle\int_0^{\dfrac{\pi}{2}}\sf \: dx \:$

$\rm :\longmapsto\:2I_1 = I_2 - log2 \bigg(x \bigg)_0^{\dfrac{\pi}{2}}\: \:$

where,

$\red{\rm :\longmapsto\:I_2 \: = \: \displaystyle\int_0^{\dfrac{\pi}{2} }\sf log(sin2x) \: dx \: - - - (2) }$

$\bf :\longmapsto\:2I_1 = I_2 -\dfrac{\pi}{2} log2 - - - - (3)$

Now, Consider Equation (2), we have

$\red{\rm :\longmapsto\:I_2 \: = \: \displaystyle\int_0^{\dfrac{\pi}{2} }\sf log(sin2x) \: dx \: }$

$\rm :\longmapsto\:Put \: 2x = y$

$\rm :\implies\:2dx = dy$

$\rm :\implies\:dx = \dfrac{dy}{2}$

When we substitute in definite integrals, the limits also changes according to the Substitution.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf \dfrac{\pi}{2} & \sf \pi \end{array}} \\ \end{gathered}$

So, on substituting all these values, we get

$\rm :\longmapsto\:I_2 = \displaystyle\int_0^{\pi}\sf logsiny \: \dfrac{dy}{2}$

$\rm :\longmapsto\:I_2 =\dfrac{1}{2} \displaystyle\int_0^{\pi}\sf logsiny \: dy$

We know,

$\boxed{ \bf{ \: \displaystyle\int_0^{a}\sf \: f(x) \: dx \: = \: \displaystyle\int_0^{a}\sf f(a - x) \: dx}}$

So using this property, we get

$\rm :\longmapsto\:I_2 =\dfrac{1}{2} \displaystyle\int_0^{\pi}\sf logsin(\pi - y) \: dy$

$\rm :\longmapsto\:I_2 =\dfrac{1}{2} \displaystyle\int_0^{\pi}\sf logsiny \: dy$

We know,

$\boxed{ \bf{ \: \displaystyle\int_0^{2a}\sf f(x) dx= 2\displaystyle\int_0^{a}\sf f(x) \: dx \: \: if \: f(2a - x) = f(x)}}$

So, using this property we get

$\rm :\longmapsto\:I_2 =2 \times \dfrac{1}{2} \displaystyle\int_0^{\dfrac{\pi}{2}}\sf logsiny \: dy$

$\rm :\longmapsto\:I_2 = \displaystyle\int_0^{\dfrac{\pi}{2}}\sf logsiny \: dy$

We know that

$\boxed{ \bf{ \: \displaystyle\int_a^{b}\sf \: f(x) \: dx \: = \: \displaystyle\int_a^{b}\sf \: f(y) \: dy}}$

$\rm :\longmapsto\:I_2 \: = \: \displaystyle\int_0^{\dfrac{\pi}{2} }\sf log(sinx) \: dx \:$

From equation (2), we concluded

$\bf\implies \:I_2 = I_1 - - - (4)$

On substituting equation (4) in equation (3), we get

$\bf :\longmapsto\:2I_1 = I_1 -\dfrac{\pi}{2} log2$

$\bf\implies \:\:I_1 = - \: \dfrac{\pi}{2} \: log2$

Hence,

$\purple{\boxed{ \bf{ \: \rm :\longmapsto\:\displaystyle\int_0^{\dfrac{\pi}{2} }\bf log(cosx) \: dx \: = - \: \dfrac{\pi}{2} \: log2 \: \: \: \: \: \: }}}$