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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals5 Marks
Question
By using the properties of definite integral, evaluate the integral $\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$
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Answer

According to the question , $ I=\int _ { 0 } ^ { \pi / 2 } ( 2 \log | \sin x | - \log | \sin 2 x | ) d x$
From $0 \ to \ {\pi\over2} , sinx \ and \ cosx$ are positive. so, $|sinx| = sinx , |cosx| = cosx$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } [ 2 \log ( \sin x ) - \log ( 2 \sin x \cos x ) ] d x$$[\because sin2x = 2sinx cosx]$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } [ 2 \log ( \sin x ) - ( \log 2 + \log ( \sin x )$$ + \log ( \cos x ) ) ] d x$$[\because log(ABC)= logA+logB+logC]$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } [ 2 \log ( \sin x ) - \log 2 - \log ( \sin x )$$ - \log ( \cos x ) ) ] d x$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } ( \log ( \sin x ) - \log 2 - \log ( \cos x ) ) d x$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } \log ( \sin x ) d x - \int _ { 0 } ^ { \pi / 2 } \log 2 d x$$ - \int _ { 0 } ^ { \pi / 2 } \log ( \cos x ) d x$
$ \Rightarrow I = \int _ { 0 } ^ { \pi / 2 } \log \sin \left( \frac { \pi } { 2 } - x \right) d x - \log 2 [ x ] _ { 0 } ^ { \pi / 2 }$$ - \int _ { 0 } ^ { \pi / 2 } \log \cos x d x$ $[\because \int_b^axdx= \int_b^a(a+b-x)dx]$
$ \Rightarrow \quad I = \int _ { 0 } ^ { \pi / 2 } \log \cos x d x - \log 2 \left[ \frac { \pi } { 2 } - 0 \right]$$ - \int _ { 0 } ^ { \pi / 2 } \log \cos x d x$
$ \therefore \quad I = - \frac { \pi } { 2 } \log 2$

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