Mathematics

# $\displaystyle \int^{\pi/2}_{0}In (\sin x)dx$ equals

$-\dfrac{\pi }{2}\ln 2$

##### SOLUTION

Consider the given integral.

$I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx}$                   …….. (1)

We know that

$\int_{a}^{b}{f\left( x \right)dx=}\int_{a}^{b}{f\left( a+b-x \right)dx}$

Therefore,

$I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin \left( \dfrac{\pi }{2}-x \right) \right)dx}$

$I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \cos x \right)dx}$                            ……. (2)

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

$2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)+\ln \left( \cos x \right)dx}$

$2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x\cos x \right)dx}$

$2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \dfrac{2\sin x\cos x}{2} \right)dx}$

$2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin 2x \right)dx}-\int_{0}^{\dfrac{\pi }{2}}{\ln \left( 2 \right)dx}$

$2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin 2x \right)dx}-\ln 2\left( x \right)_{0}^{\dfrac{\pi }{2}}$

$2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin 2x \right)dx}-\dfrac{\pi }{2}\ln 2$

Let $t=2x$

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

Therefore,

$2I=\dfrac{1}{2}\int_{0}^{\pi }{\ln \left( \sin t \right)dt}-\dfrac{\pi }{2}\ln 2$

$2I=\dfrac{2}{2}\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin t \right)dt}-\dfrac{\pi }{2}\ln 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \int_{0}^{a}{f\left( x \right)dx=2\int_{0}^{a/2}{f\left( x \right)dx}} \right]$

$2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin t \right)dt}-\dfrac{\pi }{2}\ln 2$

$2I=I-\dfrac{\pi }{2}\ln 2$

$I=-\dfrac{\pi }{2}\ln 2$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Evaluate the given integral: $\displaystyle \int_{0}^{1} (1+x)^5 dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Solve$\displaystyle \int_{0}^{\pi /2}\left ( 2\log \sin x-\log \sin 2x \right )dx$
• A. $\dfrac {\pi}{2}\log 2$
• B. $\dfrac {\pi}{4}\log 2$
• C. None
• D. $-\dfrac {\pi}{2}\log 2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of $\displaystyle\int{\left(\frac{1}{\sqrt[3]{x}+\sqrt[4]{x}}+\frac{\log{(1+\sqrt[6]{x})}}{\sqrt[3]{x}+\sqrt{x}}\right)dx}$, is equal to $\dots\dots$ where $t=x^{\tfrac{1}{12}}$ and $\theta=\log{(1+x^{\tfrac{1}{6}})}$
• A. $\displaystyle 6\left\{\sum_{r=1}^8{^8C_r{(-1)}^r\frac{t^r}{r}}+\log{(1+t)}\right\}+12\left\{\left(\frac{\theta}{3}-\frac{1}{9}\right)e^{3\theta}-\frac{3}{2}\left(\theta-\frac{1}{2}\right)e^{2\theta}+3(\theta-1)e^{\theta}-\frac{{\theta}^2}{2}\right\}+C$
• B. $\displaystyle\left\{\sum_{r=1}^8{^8C_r{(-1)}^r\frac{t^r}{r}}+\log{(1+t)}\right\}+6\left\{\left(\frac{\theta}{3}-\frac{1}{9}\right)e^{3\theta}-\frac{3}{2}\left(\theta-\frac{1}{2}\right)e^{2\theta}+3(\theta-1)e^{\theta}-\frac{{\theta}^2}{2}\right\}+C$
• C. none of the above
• D. $\displaystyle 12\left\{\sum_{r=1}^8{^8C_r{(-1)}^r\frac{t^r}{r}}+\log{(1+t)}\right\}+6\left\{\left(\frac{\theta}{3}-\frac{1}{9}\right)e^{3\theta}-\frac{3}{2}\left(\theta-\frac{1}{2}\right)e^{2\theta}+3(\theta-1)e^{\theta}-\frac{{\theta}^2}{2}\right\}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Find the following integral.
$\displaystyle\int \dfrac{x^3-1}{x^2}dx$.

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.