Correct Answer - Option 4 :
\(\frac{\pi }{2}log2\)
Let \(I = \mathop \smallint \nolimits_0^1 \frac{{{{\sin }^{ - 1}}x}}{x}dx\)
Assume sin-1 x = θ
x = sin θ
dx = cos θ dθ
When x = 0: θ = 0
\(x = 1 \)
\(\theta = \frac{\pi }{2}\)
\(I = \mathop \smallint \nolimits_0^{\pi /2} \frac{\theta }{{\sin \theta }} \cdot \cos \theta \;d\theta = \mathop \smallint \nolimits_0^{\pi /2} \theta \cdot \cot \theta \cdot d\theta \)
Let u = θ ; v = cot θ
We have:
\(\smallint uv = u\smallint v - \smallint u'\smallint v\)
\(I = \theta \mathop \smallint \nolimits_0^{\pi /2} \cot \theta \cdot d\theta - \mathop \smallint \nolimits_0^{\pi /2} \left[ {\frac{{d\theta }}{{d\theta }} \cdot \smallint \cot \theta \cdot d\theta } \right]d\theta \)
\( = \theta \cdot \log \left| {\sin \theta } \right|_0^{\pi /2} - \mathop \smallint \nolimits_0^{\pi /2} \log \left| {\sin \theta } \right|d\theta \)
In the range of \(\left[ {0,\frac{\pi }{2}} \right]\;;\sin \theta \) is positive ⇒ |sin θ| = sin θ:
\(I = \theta \cdot \left| {\log \sin \theta } \right|_0^{\frac{\pi }{2}} - \mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta d\theta \) ---(1)
Let \(\mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta \;d\theta = {I_1}\) ---(2)
Using \(\mathop \smallint \nolimits_0^a f\left( x \right)dx = \mathop \smallint \nolimits_0^a f\left( {a - x} \right)dx\)
\({I_1} = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \log \cos \theta \;d\theta \) ---(3)
From (2) and (3), we get:
\( 2{I_1} = \mathop \smallint \nolimits_0^{\pi /2} \left( {\log \sin \theta + \log \cos \theta } \right)d\theta \)
\( = \mathop \smallint \nolimits_0^{\pi /2} \log (\sin \theta \cdot \cos \theta ) \cdot d\theta \)
\(= \mathop \smallint \nolimits_0^{\pi /2} \log \left( {\frac{{\sin 2\theta }}{2}} \right) \cdot d\theta \)
\( = \mathop \smallint \nolimits_0^{\pi /2} \left( {\log \sin 2\theta - \log 2} \right)d\theta \)
\( = \mathop \smallint \nolimits_0^{\pi /2} \log \sin 2\theta \;d\theta - \mathop \smallint \nolimits_0^{\pi /2} \log 2 \cdot d\theta \)
\( = \frac{1}{2}\mathop \smallint \nolimits_0^\pi \log \sin 2\theta \;d\left( {2\theta } \right) - \log 2\;\mathop \smallint \nolimits_0^{\pi /2} d\theta \)
Property of integral:
\(\mathop \smallint \nolimits_a^b \;F\left( x \right)dx = \mathop \smallint \nolimits_a^b f\left( t \right)dt\)
\(2{I_1} = \frac{1}{2}\mathop \smallint \nolimits_0^\pi \log \sin \theta \;d\theta - \log 2\;\left| \theta \right|_0^{\pi /2}\)
\( = \mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta \;d\theta - \log 2 \cdot \frac{\pi }{2}\)
\( = \mathop \smallint \nolimits_0^{\pi /2} \log \sin \theta \cdot d\theta - \frac{\pi }{2} \cdot \log 2\)
Using (2), we get:
\(2{I_1} = {I_1} - \frac{\pi }{2}\log 2\)
\( {I_1} = - \frac{\pi }{2}\log 2\)
Substituting valve of I1 in equation (1), we get:
\(I = \frac{\pi }{2} \cdot \log \sin \frac{\pi }{2} - 0 \cdot \log \sin \theta - \left( {\frac{{ - \pi }}{2} \cdot \log 2} \right)\;\)
\( = 0 + \frac{\pi }{2} \cdot \log 2\)
\(I = \frac{\pi }{2} \cdot \log 2\)
For \(0 \cdot \log \sin \theta \;;\;\;P = \mathop {\lim }\limits_{\theta \to 0} \theta \cdot \log \sin \theta = 0\)
