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Evaluate the integral:

Evaluate The Integral class=

Sagot :

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