7.5. Trigonometric Substitutions http://www.ck12.org

Example 3:

Evaluate∫x 2 √xdx (^2) + 1.
Solution:
From the table above, letx=tanθthendx=sec^2 θdθ.Substituting into the integral,
∫ dx
x^2 √x^2 + 1 =
∫ sec (^2) θdθ
tan^2 θ√tan^2 θ+ 1.
But since tan^2 θ+ 1 =sec^2 θ,

=

∫ sec (^2) θdθ
tan^2 θsecθ

∫ secθ
tan^2 θdθ

∫ 1

cosθ

cos^2 θ

sin^2 θdθ

=

∫

cotθcscθdθ.

Sinceddθ(cscθ) =−cotθcscθ,

∫ dx

x^2 √x^2 + 1 =

∫

cotθcscθdθ

=−cscθ+C.

Looking at the triangles above, the first triangle represents our case, witha= 1 .Sox=tanθand thus sinx=
√ x
1 +x^2 ,which gives cscθ=

√

1 +x x^2 .Substituting,

∫ dx

x^2 √x^2 + 1 =−cscθ+C

=−

√ 1 +x 2

x +C.

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