Solve the following:
 \[\int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx} \]

Hint: If f is continuous on [a,b], then the function g is defined by \[g(x) = \int\limits_a^x {f(t)dt} \],\[a \leqslant x \leqslant b\], is continuous on [a,b] and differentiable on (a,b), and \[g'(x) = f(x)\]
If f is continuous on [a,b], then \[\int\limits_a^b {f(x)dx} = F(b) - F(a)\]where F is any alternative of f, that is, a function such that F’=f
A definite integral is denoted by \[\int\limits_a^b {f(x)dx} \] which represent the area bounded by the curve \[y = f(x)\], the ordinates x=a, x=b and the x-axis.

Complete step by step answer:
Step 1: let,
\[I = \int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx} \]
Now, from trigonometry properties we know that
Sin2x=2SinxCosx
Substituting the value of sin2x, we get
\[I = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log (2\sin x\cos x)]dx} \]
Step 2: from logarithm property we know that, \[\log (a.b) = \log a + \log b\], using this property of logarithm in our solution,we get
\[I = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log (2\sin x\cos x)]dx} \]
\[I = \int\limits_0^{\dfrac{\pi }{2}} {[2\log \sin x - \log 2 - \log \sin x - \log \cos x]dx} \]
\[I = \int\limits_0^{\dfrac{\pi }{2}} {[\log \sin x - \log 2 - \log \cos x]dx} \]
Step 3: opening the closed bracket so that we can integrate each term separately
\[I = \int\limits_0^{\dfrac{\pi }{2}} {[\log \sin x - \log 2 - \log \cos x]dx} \]
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} } \]


Now let
\[{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} \]
From properties of integration we know that,
\[\int\limits_0^a {f(x)dx} = \int\limits_0^a {f(a - x)} dx\]
Using this property, we get
\[{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} \]
\[{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos (\dfrac{\pi }{2} - x)dx} \]

\[{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} \]
Now, substituting the value of I1 in I we get
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} } \]
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - } \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} \]
Step 4: cancelling the term\[\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} \], we get
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - } \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} \]
\[I = - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} \]
\[I = - \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} \] (Log2 is a constant and therefore taking log2 outside the integration)
\[I = - \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}\]
\[I = - \log 2[\dfrac{\pi }{2} - 0]\]
After further simplification, we get
\[I = \dfrac{\pi }{2}\log \dfrac{1}{2}\]
Hence, the answer is\[\int\limits_0^{\dfrac{\pi }{2}} {(2\log \sin x - \log \sin 2x)dx} = \dfrac{\pi }{2}\log \dfrac{1}{2}\]

Note: While converting \[\sin x\] to \[\cos x\] or vice-versa, always use the quadrant method for giving signs (negative or positive).
Use standard result for solving easily like\[\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx = - \dfrac{\pi }{2}} \log 2 = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} \]