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(pi)/(2)int(0)^(1)log(sin x)dx...

(pi)/(2)int_(0)^(1)log(sin x)dx

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int_(0)^(pi)log sin^(2)x dx=

int_(0)^(pi//2) log sin x dx =

int_(0)^((pi)/(2))log(sin x)dx

int_(0)^(pi)log sin^(2)x dx=

If I_(1)=int_(0)^(pi//2)log (sin x)dx and I_(2)=int_(0)^(pi//2)log (sin 2x)dx , then

int_(0)^( pi)cos2x*log(sin x)dx

int_(0)^((pi)/(2))log(sin2x)dx

Show that int_(0)^(pi//2) log (sin x) dx = - pi/2 log 2