`intsinxcosx dx`

If u and v are two differentiable functions of x and int uv dx=phi(x)-int[(du)/(dx)intv dx]dx , then the value of phi(x) is-

int d/dx(f(x)) dx=d/dx int f(x)dx (True/false)

If u and v are two functions of x then prove that : int uv dx = u int v dx - int [ (du)/(dx) int v dx ] dx Hence, evaluate int x e^(x) dx

Answer the equation: int(dx)/(1+cosx)dx .

Solution of y=xdy/dx+dy/dx-(dy/dx)^^2 is

Find int(dx)/(dx+b)

int (dx)/(e^(x) - 1 ) dx is