How do you integrate ∫√5−4x−x2dx using trigonometric substitution?

1 Answer
mason m
Sep 6, 2016

(x+2)√5−4x−x2+9arcsin(x+23)2+C

Explanation:

We have:

I=∫√5−4x−x2dx=∫√−(x+2)2+9dx

From here, let 3sinθ=x+2. Thus, 3cosθdθ=dx. Plugging these in:

I=3∫cosθ√−9sin2θ+9dθ

Factoring a √9=3 from the square root:

I=9∫cosθ√−sin2θ+1dθ

Recall that since sin2θ+cos2θ=1, we know that √−sin2θ+1=cosθ.

I=9∫cos2θdθ

Recall that cos2θ=2cos2θ−1. Thus, cos2θ=12(cos2θ+1).

I=92∫(cos2θ+1)dθ=92∫cos2θdθ+92∫dθ

The first integral can be approached with substitution, or just the reverse chain rule. The second is simple:

I=94∫2cos2θdθ+92θ=94sin2θ+92θ

Using sin2θ=2sinθcosθ:

I=92sinθcosθ+92θ

Writing all in terms of sine:

I=92sinθ√1−sin2θ+92θ

Recall that sinθ=x+23, so we also see that θ=arcsin(x+23):

I=92(x+23)√1−(x+23)2+92arcsin(x+23)

I=3(x+2)2√9−(x+2)29+92arcsin(x+23)

The √19=13 will come out of the square root and cancel with the 3:

I=x+22√9−(x+2)2+92arcsin(x+23)

I=(x+2)√5−4x−x2+9arcsin(x+23)2+C

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