Let, sin x = t
Differentiating both side with respect to x
\(\cfrac{dt}{dx}=cos\,x\)
⇒dt = cos x dx
At x = 0, t = 0
At x = π\2, t = 1

By using the concept of partial fraction

1 = A(1 + t) + B(2 + t)
1 = (A + 2B) + t(A + B)
A + 2B = 1, A + B = 0
A = -1, B = 1
