Find ∫dx(x2+9)3/2.
Solution The integral contains a square root (x2+9)3/2=(√x2+9)3 that is of the form √x2+a2, where a=3. We use the substitution x=3tanθ, −π2<θ<π2. Then dx=3sec2θdθ. Since (x2+9)3/2=↑x=3tanθ(9tan2θ+9)3/2=93/2(tan2θ+1)3/2=↑tan2θ+1=sec2θ27(sec2θ)3/2=↑secθ>027sec3θ
we have ∫dx(x2+9)3/2=∫3sec2θdθ27sec3θ=19∫dθsecθ=19∫cosθdθ=19sinθ+C
To express the solution in terms of x, use either the right triangles in Figure 5 or identities.
From the right triangles, the hypotenuse is √x2+32=√x2+9. So, sinθ=x√x2+9. Then ∫dx(x2+9)3/2=19sinθ+C=x9√x2+9+C