`int_0^1y` dx where `x=cos2y`

int_(0)^(1)ydx where x=cos2y

int_(0)^(a)y^(2)dx where x=acosthetaandy=bsintheta

int_(a)^(2a)xy dx where x^(2)-y^(2)=a^(2) and y>=0

If the area bounded by the curve |x+y|+|x-y|=2 is A, then the value of int_(0)^(6A) {x}dx= (where {x}, denotes the fractional part function.

(dy)/(dx)=y+int_(0)^(1)ydx given y=1 where x=0

Evaluate (i) int_0^pi (x sin x)/(1+cos^2 x) dx Evaluate (ii) int_0^pi (4x sin x)/(1+ cos^2 x) dx

cos^(2)(x-2y)=1-2(dy)/(dx), where x-2y=u