Unique limits of integration

Calculus Level 3

arccos ( ln ( π / 4 ) ) arccos ( ln ( π / 2 ) ) ( sin x ) e cos x cot ( e cos x ) d x \large \displaystyle \int \limits_{\arccos\left( \ln \left(\pi/4 \right) \right)}^{\arccos \left( \ln \left(\pi/2 \right) \right)} (-\sin x)e^{\cos x} \cot \left( e^{\cos x}\right) \, dx

If the above integral can be represented in the form ln a \ln \sqrt{a} , find a a .

Clarification : e 2.71828 e \approx 2.71828 denotes the Euler's number .


The answer is 2.

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1 solution

Rishabh Jain
Apr 21, 2016

Put e cos x = u \large e^{\cos x}=u

\implies e cos x ( sin x ) d x = d u \large e^{\cos x}(-\sin x)\mathrm{d}x=\mathrm{d}u

and lower and upper limits of integration are

e cos ( cos 1 ( ln ( π / 4 ) ) ) \color{#D61F06}{\Large e^{\cos (\color{#20A900}{\cos^{-1}\left( \ln \left(\pi/4 \right) \right)})}}

and e cos ( cos 1 ( ln ( π / 2 ) ) ) \large \color{#3D99F6}{\Large e^{\cos(\color{#20A900}{\cos^{-1}\left( \ln \left(\pi/2 \right) \right)})}}

i.e π 4 \large\color{#D61F06}{\dfrac{\pi}{4}} and π 2 \large\color{#3D99F6}{\dfrac{\pi}{2}} respectively.

T = π / 4 π / 2 cot u d u \Large\mathfrak{T}=\displaystyle\int_{\color{#D61F06}{\pi/4}}^{\color{#3D99F6}{\pi/2}}\cot u ~\mathrm{d}u

= ln ( sin u ) π / 4 π / 2 \LARGE =\left|\ln(\sin u)\right|_{\color{#D61F06}{\pi/4}}^{\color{#3D99F6}{\pi/2}} = ln ( 2 ) \LARGE =\ln(\sqrt 2)

a = 2 \Huge \therefore \boxed{\color{#69047E}{a=2}}

In the third last step , shouldn't it be ln(sinu) ?

Ankit Kumar Jain - 4 years, 1 month ago

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Correct......

Rishabh Jain - 4 years, 1 month ago

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