Question

evaluate the following definite integrals​

Answer 1

Question :-

Evaluate the following integral

$\rm \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(cosx) \: dx$

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

Given integral is

$\rm \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(cosx) \: dx$

Let assume that

$\rm \: I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(cosx) \: dx - - - (1)$

We know,

$\boxed{ \rm{ \:\displaystyle\int_{0}^{a}f(x)dx \: = \: \displaystyle\int_{0}^{a}f(a - x)dx \: \: }}$

So, using this property, we have

$\rm \: I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: logcos\bigg(\dfrac{\pi}{2} - x\bigg) \: dx$

$\rm \: I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(sinx) \: dx - - - (2)$

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

$\rm \:2 I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \:[ log(cosx) + log(sinx) ]\: dx$

$\rm \: 2I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \:log(sinx \: cosx) \: dx$

$\rm \: 2I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \:log\bigg(\dfrac{2 \: sinx \: cosx}{2} \bigg) \: dx$

$\rm \:2 I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \:log\bigg(\dfrac{sin2x}{2} \bigg) \: dx$

$\rm \:2I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: [log(sin2x) - log2] \: dx$

$\rm \:2I = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \:log(sin2x) \: dx - log2\displaystyle\int_{0}^{\dfrac{\pi}{2}} dx$

$\rm \:2I = I_1 - log2\bigg[x\bigg]_{0}^{\dfrac{\pi}{2}}$

where,

$\rm \: I_1 = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(sin2x) \: dx - - - (3)$

$\rm \:2 I = I_1 - log2\bigg(\dfrac{\pi}{2} - 0 \bigg)$

$\rm\implies \:2I = I_1 - \dfrac{\pi}{2}log2 - - - (4)$

Now, Consider

$\rm \: I_1 = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(sin2x) \: dx$

can be rewritten as

$\rm \: I_1 = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: logsin2\bigg(\dfrac{\pi}{2} - x\bigg) \: dx$

$\rm \: I_1 = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: logsin\bigg(\pi -2 x\bigg) \: dx$

$\rm \: I_1 = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(sin2x) \: dx$

So, above can be rewritten as

$\rm \: I_1 = 2\displaystyle\int_{0}^{ \dfrac{\pi}{4} } \: log(sin2x) \: dx$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm \: 2x = y$

$\rm \: 2dx = dy$

Now, in definite integrals, when we substitute, we have to change the limits too.

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf \dfrac{\pi}{4} & \sf \dfrac{\pi}{2} \end{array}} \\ \end{gathered}$

So, above integral can be rewritten as

$\rm \: I_1 = \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(siny) \: dy$

Now, using equation (2), we get

$\rm\implies \:I_1 = I - - - (5)$

On substituting the value in equation (4), we get

$\rm\implies \:2I = I - \dfrac{\pi}{2}log2$

$\rm\implies \:I = - \dfrac{\pi}{2}log2$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} } \: log(cosx) \: dx \: = \: - \: \dfrac{\pi}{2} \: log2 \: \: }}$

$\rule{190pt}{2pt}$

Properties Used :-

$\boxed{ \rm{ \:\displaystyle\int_{0}^{a}f(x)dx \: = \: \displaystyle\int_{0}^{a}f(a - x)dx \: \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b}f(x)dx \: = \: \displaystyle\int_{a}^{b}f(y)dy \: \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{2a}f(x)dx \: = \: 2\displaystyle\int_{0}^{a}f(x)dx \: \: \: if \: f(2a - x) = f(x) \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b}f(x)dx \: = \: \displaystyle\int_{a}^{b}f(a + b - x)dx \: \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{2a}f(x)dx \: = 0 \: \: \: if \: f(2a - x) = - f(x) \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^{a}f(x)dx \: = 0 \: \: \: if \: f( - x) = - f(x) \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^{a}f(x)dx \: = \: 2\displaystyle\int_{0}^{a}f(x)dx \: \: \: if \: f(- x) = f(x) \: }}$