If I=∫0π/2sinxloge(sin x )dx and . Then the value of ‘I’ is

I=12∫0π/2sinx(In(sin2x))dx,put cos x = t I= 12∫01In(1−t2)dt=12 ∫01(−t2−(−t2)22+(−t2)33+....) dt = −12[13+110+121+....|]= −[16+120+142+.....]=−[(12−13)+(14−15)+....]=loge2−1And A=154+In2

If I=∫0π/2sinxloge(sin x )dx and . Then the value of ‘I’ is

I=12∫0π/2sinx(In(sin2x))dx,put cos x = t I= 12∫01In(1−t2)dt=12 ∫01(−t2−(−t2)22+(−t2)33+....) dt = −12[13+110+121+....|]= −[16+120+142+.....]=−[(12−13)+(14−15)+....]=loge2−1And A=154+In2

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If $I= \int_{0}^{π/2} {sinxlog}_{e} (sin x) dx and$ . Then the value of ‘I’ is

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$\log_{e} 2 + 1$

$\log_{e} 4$

$\log_{e} 2 - 1$

$\log_{e} 4 - 1$

answer is A.

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Detailed Solution

$I= \frac{1}{2} \int_{0}^{π/2} sinx (In ({sin}^{2} x)) dx,put cos x = t$
I= $\frac{1}{2} \int_{0}^{1} In (1 - t^{2}) dt= \frac{1}{2} \int_{0}^{1} (- t^{2} - \frac{{(- t^{2})}^{2}}{2} + \frac{{(- t^{2})}^{3}}{3} + ....)$ dt = $- \frac{1}{2} [\frac{1}{3} + \frac{1}{10} + \frac{1}{21} + .... |]$
= $- [\frac{1}{6} + \frac{1}{20} + \frac{1}{42} + .....] = - [(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{5}) + ....] = \log_{e} 2 - 1$
And $A= \frac{15}{4} +In2$