Question

$\huge\displaystyle \red{\tt\int_{0}^{ \frac{\pi}{2} } ln( \sin(x) ) \: dx}$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \tt{I = \int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \right) \: dx }$

$\displaystyle \tt{ \implies \: I = \int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin \left( \dfrac{\pi}{2} - x \right) \right) \: dx }$

$\displaystyle \tt{ \implies \: I = \int^{ \frac{\pi}{2} }_{0} \: \ln \left( cos(x) \right) \: dx }$

$\displaystyle \tt{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \right) \: dx + \int^{ \frac{\pi}{2} }_{0} \: \ln \left( cos(x) \right) \: dx }$

$\displaystyle \tt{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \right) +\ln \left( cos(x) \right) \: dx }$

$\displaystyle \tt{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(x) \: cos(x) \right) \: dx }$

$\displaystyle \tt{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( \dfrac{sin(2x)}{2} \right) \: dx }$

$\displaystyle \tt{ \implies \: 2I =\int^{ \frac{\pi}{2} }_{0} \: \ln \left( sin(2x) \right) \: dx - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}$

Put 2x = t, so,

$\displaystyle \tt{ \implies \: 2I = \dfrac{1}{2} \int^{ \pi}_{0} \: \ln \left( sin(t) \right) \: dt - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}$

$\displaystyle \tt{ \implies \: 2I = \dfrac{1}{2} \cdot2 \int^{ \frac{\pi}{2}}_{0} \: \ln \left( sin(t) \right) \: dt - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}$

$\displaystyle \tt{ \implies \: 2I = \int^{ \frac{\pi}{2}}_{0} \: \ln \left( sin(t) \right) \: dt - \ln(2) \int^{ \frac{\pi}{2} }_{0} \: dx}$

$\displaystyle \tt{ \implies \: 2I = I - \ln(2) \left[x \right] ^{ \frac{\pi}{2} }_{0} }$

$\displaystyle \tt{ \implies \: I = - \ln(2) \left[ \dfrac{\pi}{2} - 0 \right] }$

$\displaystyle \tt{ \implies \: I = - \dfrac{\pi}{2} \ln(2) }$