Question

Prove that integration of logsinxdx limits 0 to pi/2 = integration of logcosx dx limits 0 to pi/2 = - pi/2 log2

Answer 1

Step-by-step explanation:

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Answer 2

$\huge\mathcal\green{Question:}$

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Prove that :

$\int\limits_{0}^{\pi /2} log(\sin x) \, dx = \int\limits_{0}^{\pi /2} log(\cos x) \, dx =\left [ - \dfrac{\pi}{2} (log2) \right ]$

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Let us assume

$\implies I = \int\limits_{0}^{\pi /2} log(\sin x) \, dx$ __________ ( 1 )

We can also write the expression as

$\implies I = \int\limits_{0}^{\pi /2} log \: \sin \left ( \dfrac{\pi}{2} - x \right ) \, dx$

Using Trigonometric Identity

$\fbox{\sin \left ( \dfrac{\pi}{2} - x \right ) = \cos x }$

$\implies I = \int\limits_{0}^{\pi /2} log(\cos x) \, dx$ __________( 2 )

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Adding equation ( 1 ) and ( 2 ), we get

$\implies 2I = \int\limits_{0}^{\pi /2} log(\sin x) \, dx + \int\limits_{0}^{\pi /2} log(\cos x) \, dx$

$\implies 2I = \int\limits_{0}^{\pi /2} (log\: \sin x + log \: cos x)\, dx$

Using Logarithmic Identity

$\fbox{log(a)+log(b) = log(ab) }$

$\implies 2I = \int\limits_{0}^{\pi /2} log (\sin x \: \cos x ) \, dx$

Multiplying and Dividing by 2 on RHS

$\implies 2I = \int\limits_{0}^{\pi /2} log \left ( \dfrac{2 \sin x \: \cos x}{2} \right ) \, dx$

Using Trigonometric Identity

$\fbox{2 \: \sin x \: \cos x = \sin 2x}$

$\implies 2I = \int\limits_{0}^{\pi /2} log \left ( \dfrac{\sin 2x}{2} \right ) \, dx$

$\fbox{log \left ( \dfrac{m}{n} \right ) = log(m) - log(n) }$

$\implies 2I = \int\limits_{0}^{\pi /2} log ( \sin 2x - log2) \, dx$

Separating the integral in both terms

$\implies 2I = \int\limits_{0}^{\pi /2} log(\sin 2x)dx - \int\limits_{0}^{\pi /2} log2 \, dx$

$\implies 2I = \int\limits_{0}^{\pi /2} log(\sin 2x) dx - (log2)\int\limits_{0}^{\pi /2}dx$

Since we know that

$\fbox{\int dx = x}$

$\implies 2I = \underbrace{\int\limits_{0}^{\pi /2} log(\sin 2x)dx}_{A} - \dfrac{\pi}{2} (log2)$

$\implies 2I = A - \dfrac{\pi}{2} (log2)$ _________ ( 3 )

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Finding the value of A

$A=\int\limits_{0}^{\pi /2} log(\sin 2x)dx$ ________ ( 4 )

Putting $2x=t$

$\implies 2x = t$

Differentiating Both sides w.r.t x

$\implies 2 \dfrac{dx}{dx} = \dfrac{dt}{dx}$

$\implies dx = \dfrac{dt}{2}$ ______ ( 5 )

From equation ( 4 ) and equation ( 5 )

$\implies A = \int\limits_{0}^{\pi} log(\sin t)\dfrac{dt}{2}$

Using Property of Integration

$\fbox{\int\limits_{0}^{a} k\: f(x)\, dx= k \: \: \int\limits_{0}^{a} f(x) \, dx}$

$\implies A = \dfrac{1}{2} \int\limits_{0}^{\pi} log(\sin t)dt$

Using Property of integration

$\fbox{\int\limits_{0}^{2a} f(x) \, dx =2 \: \int\limits_{0}^{a} f(x) \, dx}$

$\implies A = \dfrac{1}{2} \times 2 \int\limits_{0}^{\pi /2} log(\sin t) dt$

Using of integration

$\fbox{\int\limits_{a}^{b} f(x) \, dx = \int\limits_{a}^{b} f(t) \, dt }$

$\implies A = \int\limits_{0}^{\pi /2} log(\sin x)dx$

$\implies A = I$ [ From equation 1 ]

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Putting A = I in equation ( 3 )

$\implies 2I = I - \dfrac{\pi}{2}(log2)$

$\implies 2I - I = \left [ - \dfrac{\pi}{2} (log2) \right ]$

$I = \left [ - \dfrac{\pi}{2} (log2) \right ]$

Hence it is proved that

$\fbox{\int\limits_{0}^{\pi /2} log(\sin x) \, dx = \int\limits_{0}^{\pi /2} log(\cos x) \, dx =\left [ - \dfrac{\pi}{2} (log2) \right ] }$

$\huge\mathcal\red{HENCE \: \: PROVED :}$