# Answer on Calculus Question for sakura

Question #19914

antiderivative of sqrt(5 + 4x - x^2)

using chain rule or integration by parts.

Can i solve it not using trigonometry?

using chain rule or integration by parts.

Can i solve it not using trigonometry?

Expert's answer

Possible intermediate steps:

integralsqrt(-x^2+4 x+5) dx

For the integrand sqrt(-x^2+4 x+5), complete the square:

= integral sqrt(9-(x-2)^2) dx

For the integrand sqrt(9-(x-2)^2), substitute u = x-2and du = dx:

= integral sqrt(9-u^2) du

For the integrand sqrt(9-u^2), substitute u = 3 sin(s)and du = 3 cos(s) ds. Then sqrt(9-u^2) = sqrt(9-9 sin^2(s)) = 3cos(s) and s = sin^(-1)(u/3):

= 9 integralcos^2(s) ds

Write cos^2(s) as 1/2 cos(2 s)+1/2:

= 9 integral (1/2cos(2 s)+1/2) ds

Integrate the sum term by term and factor out constants:

= 9 integral 1/2ds+9/2 integral cos(2 s) ds For theintegrand cos(2 s), substitute p = 2 s and dp = 2 ds:

= 9/4 integral cos(p) dp+9 integral 1/2 ds Theintegral of cos(p) is sin(p):

= (9 sin(p))/4+9integral 1/2 ds

The integral of 1/2 is s/2:

= (9 sin(p))/4+(9s)/2+constant

Substitute back for p = 2 s:

= (9 s)/2+9/4sin(2 s)+constant

Substitute back for s = sin^(-1)(u/3):

= 1/2 usqrt(9-u^2)+9/2 sin^(-1)(u/3)+constant Substitute back for u = x-2:

= 1/2 (x-2)sqrt(-x^2+4 x+5)+9/2 sin^(-1)((x-2)/3)+constant Factor the answer a different

way:

Answer: |

| = 1/2 ((x-2) sqrt(-x^2+4 x+5)+9sin^(-1)((x-2)/3))+constant

integralsqrt(-x^2+4 x+5) dx

For the integrand sqrt(-x^2+4 x+5), complete the square:

= integral sqrt(9-(x-2)^2) dx

For the integrand sqrt(9-(x-2)^2), substitute u = x-2and du = dx:

= integral sqrt(9-u^2) du

For the integrand sqrt(9-u^2), substitute u = 3 sin(s)and du = 3 cos(s) ds. Then sqrt(9-u^2) = sqrt(9-9 sin^2(s)) = 3cos(s) and s = sin^(-1)(u/3):

= 9 integralcos^2(s) ds

Write cos^2(s) as 1/2 cos(2 s)+1/2:

= 9 integral (1/2cos(2 s)+1/2) ds

Integrate the sum term by term and factor out constants:

= 9 integral 1/2ds+9/2 integral cos(2 s) ds For theintegrand cos(2 s), substitute p = 2 s and dp = 2 ds:

= 9/4 integral cos(p) dp+9 integral 1/2 ds Theintegral of cos(p) is sin(p):

= (9 sin(p))/4+9integral 1/2 ds

The integral of 1/2 is s/2:

= (9 sin(p))/4+(9s)/2+constant

Substitute back for p = 2 s:

= (9 s)/2+9/4sin(2 s)+constant

Substitute back for s = sin^(-1)(u/3):

= 1/2 usqrt(9-u^2)+9/2 sin^(-1)(u/3)+constant Substitute back for u = x-2:

= 1/2 (x-2)sqrt(-x^2+4 x+5)+9/2 sin^(-1)((x-2)/3)+constant Factor the answer a different

way:

Answer: |

| = 1/2 ((x-2) sqrt(-x^2+4 x+5)+9sin^(-1)((x-2)/3))+constant

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