Rinkoo Gupta
Last Activity: 10 Years ago
Integral 0 to pi (log sinx) dx
=2 integral lim 0 to pi/2 ( log sinx) dx using property of definite integral
Now let I= integral lim 0 to pi/2 log sinx dx eq no (1)
=integral lim 0 to pi/2 log sin(pi/2-x) dx
=integral lim 0 to pi/e log cosx dx eq no(2)
Adding eq (1) and (2) we get
2I=integral lom 0 to pi/2 (log sinx + log cosx) dx
=> 2I=integral lim 0 to pi/2 log sinx.cosx dx
=> 2I=integral lim 0 to pi/2 log (sinxcosx) dx
=> 2I=integral lim 0 to pi/2 log ( 2sinxcosx/2) dx
=> 2I=integral 0 to pi/2 log sin2x dx - integral 0 to pi/2 log2 dx
let 2x=t in first integral => 2dx=dt and new lim are 0 to pi
=> 2I=1/2integral 0 to pi log sint dt -pi/2log2
=>2I=1/2 .2 integral 0 to pi/2 log sinx dx -pi/2 log 2 using properties of definite integral
=> 2I=integral 0 to pi/2 log sinx dx -pi/2 log2
=>2I=I -pi/2 log 2
=>I=-pi/2 log 2
Hence The value of integral 0 to pi (log sinx) dx=2(-pi/2 log 2)=-pi log2 Ans.
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Rinkoo Gupta
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