Integration + Limit

Calculus Level 4

lim a π 2 0 a cos x ln ( cos x ) d x \large\lim_{a\to \frac\pi2 ^-} \int_0^a \cos x \ln(\cos x) \, dx

If the integral above can be expressed as ln c b \ln c - b , where b b and c c are positive integers, find b + c b +c .

2 1 5 4 3

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1 solution

Rishabh Jain
Mar 28, 2016

Applying IBP:

sin x ln cos x 0 a + 0 a sin 2 x d x cos x \sin x\ln \cos x|_0^a+\int_0^a \dfrac{\sin^2 x dx}{\cos x} = sin a ln cos a + 0 a ( 1 cos 2 x ) d x cos x = \sin a\ln \cos a +\int_0^a \dfrac{(1-\cos^2 x) dx}{\cos x}

= sin a ln cos a + ln ( sec a + tan a ) sin a =\sin a\ln \cos a+\ln (\sec a+\tan a)-\sin a

= ln ( ( cos a ) sin a 1 ( 1 + sin a ) ) sin a =\ln ((\cos a)^{\sin a-1}(1+\sin a) )-\sin a

Applying limits we get

lim a π 2 ( ln ( cos a ) sin a 1 ( 1 + sin a ) ) 1 \lim_{a\to \frac{\pi}{2}^-} (\ln (\cos a)^{\sin a-1}(1+\sin a))-1

= lim a π 2 ( ln ( cos a ) sin a 1 ) + ln 2 1 = ln 2 1 =\lim_{a\to \frac{\pi}{2}^-} (\ln (\cos a)^{\sin a-1} )+\ln 2-1=\ln 2-1

Hence 2 + 1 = 3 \Large 2+1=\boxed 3 .

Thanks for the solution.

Rishabh Deep Singh - 5 years, 2 months ago

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My pleasure...... :-)

Rishabh Jain - 5 years, 2 months ago

Rishabh in what standard you syudy now

Patel Akshay - 4 years, 3 months ago

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