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EXAMPLE 2Finding an Integral Containing x2+9

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

Figure 5 tanθ=x3, π2<θ<π2

we have dx(x2+9)3/2=3sec2θdθ27sec3θ=19dθsecθ=19cosθdθ=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=xx2+9. Then dx(x2+9)3/2=19sinθ+C=x9x2+9+C