Using integral `int_0^(-pi/2)ln(sinx)dx=-int_0^piln(secx)dx=-pi/2ln2 and int_0^(pi/2)ln(tanx)
140 views · Published 13 October 2018 · 3:19 · Indexed 7 October 2026
Channel: Doubtnut · 2018 · Education
To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW Using integral `int_0^(-pi/2)ln(sinx)dx=-int_0^piln(secx)dx=-pi/2ln2 and int_0^(pi/2)ln(tanx)dx=0 and int_0^(pi/4)ln(1+tanx)dx=pi/8ln2`
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