int_(0)^((pi)/(2))(dx)/(3+2cos x)=(2)/(sqrt(5))tan^(-1)(1)/(sqrt(5))

int_(0)^((pi)/(2))(dx)/(2cosx+4sinx)=(1)/(sqrt(5))log((3+sqrt(5))/(2))

int_(0)^((3pi)/(4))(sinxdx)/(1+cos^(2)x)=(pi)/(4)+tan^(-1)((1)/(sqrt(2)))

A: int (1)/(3+2 cos x)dx=(2)/(sqrt(5))"Tan"^(-1)((1)/(sqrt(5))"tan" (x)/(2))+c R: If a gt b then int (dx)/(a+b cosx)=(2)/(sqrt(a^(2)-b^(2)))Tan^(-1)[(sqrt(a-b))/(a+b)"tan"(x)/(2)]+c

int_(0)^((pi)/(2))(cos^(2)x)/(sin x+cos x)dx=(1)/(sqrt(2))(log(sqrt(2)+1))

Show that : int_(0)^((pi)/(2))(sin^(2)x)/(sin x+cos x)dx=(1)/(sqrt(2))log(sqrt(2)+1)

int_(0)^((pi)/(6))(sqrt(3 cos 2x-1))/(cos x)dx

int_(0)^((pi)/(2))(dx)/((1-2x^(2))sqrt(1-x^(2)))=(1)/(2)log(2+sqrt(3))

int_(0)^((pi)/(2))(dx)/((1-2x^(2))sqrt(1-x^(2)))=(1)/(2)log(2+sqrt(3))

If int_(0)^(1)(x+4)/(x^(2)+5)dx = a log ((6)/(5))+b tan^(-1)((1)/(sqrt(5))) , then (a, b)-=

Find the mistake of the following evaluation of the integral I=int_0^pi(dx)/(1+2sin^2x) I=int_0^pi(dx)/(cos^2x+3sin^2x) =int_0^pi(sec^2x dx)/(1+3tan^2x)=1/(sqrt(3))[tan^(-1)(sqrt(3)tanx)]pi0=0