`I`=`int_(0)^( pi)(x)/(1-cos alpha sin x)dx`

int_(0)^( pi)(dx)/(1+cos x)

int_(0)^( pi)(dx)/(1+sin x)

Evaluate int_(0)^(pi)(x dx)/(1+cos alpha sin x) ,where 0lt alpha lt pi .

int_(0)^(pi/4)(sin x+cos x)/(9+16 sin 2x)dx

If I_(1)=int_(0)^((pi)/(2))e^(sinx)(1+x cos x)dx and I_(2)=int_(0)^((pi)/(2))e^(cosx)(1-x sin x)dx, then [(I_(1))/(I_(2))] is equal to (where [x] denotes the greatest integer less than or equal to x)

If I_(n)=int_(0)^(pi)(1-sin2nx)/(1-cos2x)dx then I_(1),I_(2),I_(3),"….." are in

int_(0)^(50pi)| cos x|dx=

int_(0)^(pi//2) e^(x) (sin x + cos x) dx

Evaluate : int_(0)^(pi)(|cos x| + |sin x|)dx .

Evaluate int_(0)^(pi//2) (cos^2 x)/(1+sin x cos x)dx .