Substitute x = √7 sin(t) and dx = √7 cos(t) dt. Then
∫ √(7 - x²) dx = ∫ √(7 - (√7 sin(t))²) • √7 cos(t) dt
… = √7 ∫ √(7 - 7 sin²(t)) cos(t) dt
… = 7 ∫ √(1 - sin²(t)) cos(t) dt
… = 7 ∫ √(cos²(t)) cos(t) dt
We require that -π/2 ≤ t ≤ π/2 in order for the substitution we made to be reversible. Over this domain, cos(t) ≥ 0, so
√(cos²(t)) = |cos(t)| = cos(t)
and the integral reduces to
… = 7 ∫ cos²(t) dt
Recall the half-angle identity for cosine:
cos²(t) = (1 + cos(2t))/2
Then the integral is
… = 7/2 ∫ (1 + cos(2t)) dt
… = 7/2 (t + 1/2 sin(2t)) + C
… = 7t/2 + 7/4 sin(2t) + C
Get the antiderivative back in terms of x. Recall the double angle identity for sine:
sin(2t) = 2 sin(t) cos(t)
We have t = arcsin(x/√7), which gives
sin(t) = sin(arcsin(x/√7)) = x/√7
cos(t) = cos(arcsin(x/√7)) = √(7 - x²)/√7
Then
∫ √(7 - x²) dx = 7/2 arcsin(x/√7) + 7/4 • 2 sin(arcsin(x/√7)) cos(arcsin(x/√7)) + C
… = 7/2 arcsin(x/√7) + x/2 √(7 - x²) + C
