Solve the integral using properties of integrals .

sinx + cosx / 9 + 16sin2x dx

I = int x from 0 to pi/4 (sinx + cosx)dx / (9 + 16sin2x)

I = int x from 0 to pi/4(sinx + cosx)dx / (25 - 16(sinx - cosx)²)

Put y = sinx - cosx

or, dy = (cosx + sinx).dx

When x = 0, y = -1 and when x = pi/4 , y = 0

I = int y from -1 to 0 (dy / 25 - 16y²)

I = (1/16) int y from -1 to 0 ( dy/ (5/4)² - y² )

I = (1/16) (2/5) |log(5 + 4y / 5 - 4y)| with limit of y from -1 to 0

I = (1/40) ( 0 - log(1/9) )

I = (1/40) (log9)

Using properties of integrals we have to solve