Integral sqrt(25-9x^2) dx

Question

Need to integrate from -5/3 to 5/3. Function is sqrt(25-9x^2) dx.

Given substitution is to make u=3x.

 

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The answer is 25pi/6. I am assuming a trig identity comes into play somewhere. Help?

Answer 1

25-9x²=5²-3²x²=3²((5/3)²-x²).

Let x=(5/3)sinθ, then dx=(5/3)cosθdθ, and (5/3)²-x²=(5/3)²-(5/3)²sin²θ=(5/3)²(1-sin²θ)=((5/3)cosθ)² (trig identity).

So √(25-9x²)dx=[√(3²((5/3)cosθ)²)][(5/3)cosθdθ]=

5cosθ.(5/3)cosθdθ=(25/3)cos²θdθ.

cos(2θ)=2cos²θ-1, so cos²θ=½(1+cos(2θ)) (trig identity).

(25/3)cos²θdθ=(25/6)(1+cos(2θ))dθ. Therefore, (25/6)∫(1+cos(2θ))dθ=(25/6)(θ+sin(2θ)/2).

Now to convert and apply the limits.

When x=-5/3, -5/3=(5/3)sinθ, sinθ=-1, θ=3π/2 (lower limit); x=5/3, sinθ=1, θ=π/2 (upper limit). However, 3π/2>π/2, so the limits need to be reversed.

Plug these revised limits into the integrated expression:

(25/6)[(3π/2+sin(3π)/2)-(π/2+sin(π)/2)]=(25/6)[π]=25π/6.