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The value of \(\displaystyle\int \limits_0^{\pi /2} (\ln {e^{\sin x}}) \times {e^{\cos x}}dx\)
1. e + 1
2. π/2
3. e -1
4. 0
5. None of these

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Correct Answer - Option 3 : e -1

Concept:

Logarithmic properties:

  • ln ef(x) = f(x)

Definite Integral properties:

  • \(\displaystyle\int \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx\;}} = - {\rm{\;}}\displaystyle\int \limits_{\rm{b}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

Calculation:

\({\rm{I}} = \displaystyle\int \limits_0^{{\rm{\pi }}/2} \left( {\ln {{\rm{e}}^{\sin {\rm{x}}}}} \right) \times {{\rm{e}}^{\cos {\rm{x}}}}{\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \displaystyle\int \limits_0^{{\rm{\pi }}/2} \sin {\rm{x}} \times {{\rm{e}}^{\cos {\rm{x}}}}{\rm{dx}}\) [∵ ln ef(x) = f(x)]

Let ecos x = t

Differentiating both sides, we get

ecos x (-sin x) dx = dt

∴ ecos x sin x = -dt

x

0

π/2

t

e

1


So, \({\rm{I}} = - \displaystyle\int \limits_{\rm{e}}^1 {\rm{dt}}\)

\(\Rightarrow {\rm{I}} = \displaystyle\int \limits_1^{\rm{e}} {\rm{dt}}\)

\(\Rightarrow I = \left[ t \right]_1^e\)

⇒ I = e - 1

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