Question

integration underroot 1-sin square X. Sin2x​

Answer 1

Answer:

Write 1 = sin^2 x + cos^2 x (Identity)

And write sin2x = 2sinx cosx

So 1 + sin2x = sin^2 x + cos^2 x + 2sinx cosx

=> Square root of 1 + sin2x = Square root of (sin^2 x + cos^2 x + 2sinx cosx)

= (sinx + cosx) or (-sinx -cosx)

So integral of sinx + cos x dx = -cosx + sinx + c

and integral of -sinx - cos x dx = cosx - sinx + c

Step-by-step explanation: