Mathematics

Solve: $$\displaystyle \int^{\pi/2}_{0}(2log \sin x - log \sin 2x)dx$$


ANSWER

$$\frac{\pi}{2}log\frac{1}{2}$$


SOLUTION
Consider, $$I=$$$$\displaystyle \int^{\pi/2}_{0}(2\log \sin x - log \sin 2x)dx$$

$$I=$$$$\displaystyle \int^{\pi/2}_{0}\log\left(\dfrac{ \sin^2 x}{2\sin x.\cos x}\right)dx$$

$$I=$$$$\displaystyle \int^{\pi/2}_{0}\log \tan xdx$$$$+\displaystyle \int^{\pi/2}_{0}\log \dfrac{1}{2}dx$$$$=\displaystyle \int^{\pi/2}_{0}\log \tan xdx$$$$+\dfrac{\pi}{2}\log \dfrac{1}{2}dx$$........(1).

$$I=$$$$\displaystyle \int^{\pi/2}_{0}\log \tan \left(\dfrac{\pi}{2}-x\right)dx+\dfrac{\pi}{2}\log\dfrac{1}{2}$$ [ Using properties of definite integral]

$$I=$$$$\displaystyle \int^{\pi/2}_{0}\log  \cot xdx+\dfrac{\pi}{2}\log \dfrac{1}{2}$$ .......(2).

Now adding (1) and (2) we get,

$$2I=$$$$\displaystyle \int^{\pi/2}_{0}\log  1 dx+\pi \log \dfrac{1}{2}$$

$$2I=\pi \log \dfrac{1}{2}$$

$$I=\dfrac{\pi}{2}\log \dfrac{1}{2}$$.
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
Integrate $$\displaystyle \int \dfrac 1{\sqrt {3-x^2}}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Hard
$$\displaystyle \int(\cot x+x)\cot^{2}x.dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
Evaluate: $$\displaystyle \int \cot x dx$$. 

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate the following integrals:
$$\displaystyle \int { \tan ^{  }{ x } \sec ^{ 4 }{ x }  } dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
Evaluate $$\int \tan^{4} x\ dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer