`int_0^pi (sinx)/(1+cos^2x)dx=pi(cosalpha)/(1+sin^2alpha)`

12 views · Published 12 October 2018 · 3:48 · Indexed 26 September 2026

Channel: Doubtnut · 2018 · Education

Watch on YouTube

To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW `int_0^pi (sinx)/(1+cos^2x)dx=pi(cosalpha)/(1+sin^2alpha)`

More from this channel