Find the integral
`I=int 1/(a^2-b^2cos^2x)dx`

Integrate : int (1-sinx)/(cos^2x)dx.

Integrate I= int(sin2x-cos2x)^(2)dx

Find the integral int x^(2)(1-(1)/(x^(2)))dx

Evaluate the following integrals: int (1)/(sin^(2)x cos^(2)x) dx

Find the integrals functions int (cos2x)/sqrt(1+sin 2x) dx

Evaluate the following integrals : int(1)/(1+cos2x)dx

If the value of int_0^pi(x/(1+sinx))^2dx=lambda , then find the value of the integral =int_0^pi[(2x^2*cos^2(x/2))/(1+sinx)^2]dx

Find the integration int(sin2x cos x.dx)

Evaluate the following integral: int_0^(pi//2)1/(a^2sin^2x+b^2cos^2x)dx

The integral I=int(sin(x^(2))+2x^(2)cos(x^(2)))dx (where =xh(x)+c , C is the constant of integration). If the range of H(x) is [a, b], then the value of a+2b is equal to