Question

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

Answer 1

Correct Answer - Option 3 : e -1

Concept:

Logarithmic properties:

• ln e^f(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 e^f(x) = f(x)]

Let e^{cos x} = t

Differentiating both sides, we get

^⇒ e^{cos x} (-sin x) dx = dt

∴ e^{cos 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