JEE Integration

Calculus Level 4

Evaluate

I = 0 π / 2 x cot x d x . I = \int_{0}^{\pi/2} x \cot{x} \, dx.


The answer is 1.0888.

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2 solutions

Sandeep Rathod
Nov 5, 2014

ILATE RULE

x log ( sin x ) 0 π / 2 0 π / 2 log ( sin x ) d x = ( π 2 log 2 ) = 1.08879 x\log(\sin x)\bigg|_0^{\pi/2} - \int\limits_0^{\pi/2}\log(\sin x)\mathrm dx= -\left(-\frac{\pi}{2}\log2\right) = 1.08879

Care to elaborate on your solution? Currently as it is, the solution is very incomplete and doesn't help in understanding how the integral is evaluated.

You missed mentioning many things like your use of IBP (note that ILATE is just a mnemonic that helps when you use IBP, it isn't a rule by itself). You also need to at least give a brief explanation of how the simpler integrals and expressions are evaluated to get to the final answer.

Also, it needs to be mentioned in the solution that this is an improper definite integral which would explain why we do the calculation taking limits at endpoints.

Prasun Biswas - 5 years, 8 months ago

Surely at least one and possibly both terms in your solution are undefined (ln(0))?

Joseph Mulvey - 6 years, 7 months ago

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Take the limit lim x 0 + x log ( sin x ) = 0 \lim_{x \rightarrow 0^+} \ \ x \cdot \log (\sin x)=0

Pratik Shastri - 6 years, 7 months ago

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l o g s i n x 1 x \frac{logsinx}{\frac{1}{x}} \frac{\infty}{\infty} form , apply L.H

sandeep Rathod - 6 years, 7 months ago

I can't understand,

Kaushik Deka - 3 years, 10 months ago

its "ln" not "log"

Rishabh Jain - 6 years, 7 months ago

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using "ln" or "log" does not make difference.

Parag Zode - 6 years, 7 months ago
Jam M
Feb 21, 2021

Using integration by parts with u = x u = x and d v = cot x d x dv = \cot x dx , we have: I = x ln ( sin x ) 0 π / 2 0 π / 2 ln ( sin x ) d x I = x \ln(\sin x)|_0^{\pi/2} - \int_0^{\pi/2} \ln(\sin x) dx

Let I 1 = 0 π / 2 ln ( sin x ) d x I_1 = \int_0^{\pi/2} \ln(\sin x) dx , and note that:

I 1 = 0 π / 2 ln ( 2 sin ( x 2 ) cos ( x 2 ) d x = 0 π / 2 ln ( 2 ) d x + 0 π / 2 ln ( sin ( x 2 ) ) d x + 0 π / 2 ln ( cos ( x 2 ) ) d x = π l n ( 2 ) 2 + I 2 + I 3 I_1 = \int_0^{\pi/2} \ln(2 \sin \left( \dfrac{x}{2} \right) \cos \left(\dfrac{x}{2} \right) dx \\ = \int_0^{\pi/2} \ln(2) dx + \int_0^{\pi/2} \ln \left( \sin \left( \dfrac{x}{2} \right) \right) dx + \int_0^{\pi/2} \ln \left( \cos \left( \dfrac{x}{2} \right) \right) dx \\ = \dfrac{\pi ln(2)}{2} + I_2 + I_3

Let x = 2 u x = 2u so that I 2 + I 3 = 2 0 π / 4 ln ( sin u ) d u + 2 0 π / 4 ln ( cos u ) d u I_2 + I_3 = 2 \int_0^{\pi/4} \ln (\sin u) du + 2 \int_0^{\pi/4} \ln(\cos u) du

Let u = π 2 t u = \dfrac{\pi}{2} - t so that 2 0 π / 4 ln ( cos u ) d u = 2 π / 4 π / 2 ln ( sin t ) d t 2 \int_0^{\pi/4} \ln(\cos u) du = 2 \int_{\pi/4}^{\pi/2} \ln(\sin t) dt , and so I 1 = π l n ( 2 ) 2 + 2 0 π / 2 ln ( sin x ) d x = π l n ( 2 ) 2 + 2 I 1 I_1 = \dfrac{\pi ln(2)}{2} + 2 \int_{0}^{\pi/2} \ln(\sin x) dx = \dfrac{\pi ln(2)}{2} + 2I_1 I 1 = π l n ( 2 ) 2 I_1 = -\dfrac{\pi ln(2)}{2}

Finally, I = lim b 0 b ln ( sin b ) ( π l n ( 2 ) 2 ) = 0 + π l n ( 2 ) 2 1.088 I = \displaystyle \lim_{b \to 0} b \ln(\sin b) - \left( -\dfrac{\pi ln(2)}{2} \right) = 0 + \dfrac{\pi ln(2)}{2} \sim 1.088 , where the limit is evaluated using L’Hospital’s Rule.

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