TRIGONOMETRIC SUBSTITUITION
√ a2 - x 2 implies x = asinθ
√ a2 + x 2 implies x = atanθ
√ x 2 - a2 implies x = asecθ
Things to remember
sin2 θ + cos2 θ = 1
2 2
1 + tan θ = sec θ
Thats it
lets take some examples
x3
(a) ∫ √x 2
+9
dx
from the expression we know that
√ a2 + x 2 is just the same as √ x2 + 9
then
we let x = 3tanθ (remember √ 9 =3 )
thus dx = 3sec2 θdθ
then we substitute
x3 (3tanθ)3 27tan
3
θ
∫ √x 2
+9
dx = ∫ √ ( tanθ)
3
2
+9
(3sec θ)dθ = ∫
2
√ 9tan2 θ + 9
3sec
2
θdθ
3
= ∫ √ (tan θ θ )
9
27tan
2
+1
3sec
2
θdθ
{remember from trig identities
tan2 θ + 1 = sec2 θ }
3 3
27tan θ 27tan θ
= ∫ √ sec θ
9
2
3sec
2
θdθ = ∫ 3secθ
2
3sec θdθ
3
27tan θ
= ∫ 3secθ
3secθsecθdθ = ∫27tan θsecθdθ 3
√ a2 - x 2 implies x = asinθ
√ a2 + x 2 implies x = atanθ
√ x 2 - a2 implies x = asecθ
Things to remember
sin2 θ + cos2 θ = 1
2 2
1 + tan θ = sec θ
Thats it
lets take some examples
x3
(a) ∫ √x 2
+9
dx
from the expression we know that
√ a2 + x 2 is just the same as √ x2 + 9
then
we let x = 3tanθ (remember √ 9 =3 )
thus dx = 3sec2 θdθ
then we substitute
x3 (3tanθ)3 27tan
3
θ
∫ √x 2
+9
dx = ∫ √ ( tanθ)
3
2
+9
(3sec θ)dθ = ∫
2
√ 9tan2 θ + 9
3sec
2
θdθ
3
= ∫ √ (tan θ θ )
9
27tan
2
+1
3sec
2
θdθ
{remember from trig identities
tan2 θ + 1 = sec2 θ }
3 3
27tan θ 27tan θ
= ∫ √ sec θ
9
2
3sec
2
θdθ = ∫ 3secθ
2
3sec θdθ
3
27tan θ
= ∫ 3secθ
3secθsecθdθ = ∫27tan θsecθdθ 3
