Navigation 
 Home 
 Search 
 Site map 

 Contact Graeme 
 Home 
 Email 
 Twitter

 Skip Navigation LinksMath Help > Calculus > Integral > Trig Substitutions

Trigonometric Substitutions

sqrt(b2x2-a2)  ==>  x=(a/b) sec θ

If the integral contains sqrt(b2x2-a2), then make the substitution x=(a/b) sec θ.

bx

 sqrt(b2x2-a2)
a

Then

dx = (a/b) sec θ tan θ dθ

sec θ = bx/a

Here's an example:

sqrt(b2x2-a2)/x dx

a sqrt(b2x2/a2-1)/x dx

a sqrt(sec2 θ-1)/x dx

and, because sec2θ - 1 = tan2θ,

a sqrt(tan2 θ)/x dx

a tan θ/x dx

a tan θ/((a/b) sec θ) (a/b) sec θ tan θ dθ

a tan2 θ d θ

and now converting tan2θ back the other way -- tan2θ = sec2θ - 1,

a (sec2 θ - 1) dθ

a tan θ - a θ

referring to the triangle at the top of this section, we convert tan θ into a ratio of side lengths, and we convert θ into any one of the six inverse trig functions (cos-1 being the simplest in this case),

sqrt(b2x2-a2) - a cos-1(a/bx)

sqrt(a2-b2x2-a)  ==>  x=(a/b) sin θ

 . . . . . . 

f the integral contains sqrt(a2-b2x2), then make the substitution x=(a/b) sin θ.

a

bx 
sqrt(a2-b2x2)

Then

dx = (a/b) cos θ dθ

sin θ = bx/a

Here's an example:

x-4(a2-b2x2)-1/2 dx

x-4(1/a)(1-b2x2/a2)-1/2 dx

x-4(1/a)(1-sin2θ)-1/2 dx

x-4(1/a)(1/cos θ) dx

x-4(1/a)(1/cos θ) (a/b) cos θ dθ

x-4 (1/b) dθ

(1/b)(a/b)-4 csc4θ dθ

(1/b)(a/b)-4 csc2θ csc2θ dθ

(1/b)(a/b)-4 (1+cot2θ) csc2θ dθ

and since d(cot θ) = -csc2θ dθ, this integral evaluates to

-(1/b)(a/b)-4 (cot θ + (1/3)cot3θ)

referring to the triangle at the top of this section, we convert tan θ into a ratio of side lengths, and we convert θ into any one of the six inverse trig functions (cos-1 being the simplest in this case),

-(1/b)(a/b)-4 (sqrt(a2-b2x2)/bx + (1/3)(sqrt(a2-b2x2)/bx)3)

-a-4b3 (sqrt(a2-b2x2)/bx + (1/3) sqrt(a2-b2x2)3/(b3x3))

-a-4 (b2sqrt(a2-b2x2)/x + (1/3) (a2-b2x2) sqrt(a2-b2x2)/x3)

-a-4 (1/3) (3b2x2sqrt(a2-b2x2)/x3 + (a2-b2x2) sqrt(a2-b2x2)/x3)

-a-4 (1/3) x-3 (3b2x2sqrt(a2-b2x2) + (a2-b2x2) sqrt(a2-b2x2))

-a-4 (1/3) x-3 (a2+2b2x2) sqrt(a2-b2x2)

-sqrt(a2-b2x2) (a2+2b2x2) / (3a4x3)

sqrt(a2+b2x2)  ==>  x=(a/b) tan θ

 If the integral contains sqrt(a2+b2x2), then make the substitution x=(a/b) tan θ.

sqrt(a2+b2x2)

 bx 
a

Then

dx = (a/b) sec2 θ dθ

tan θ = bx/a

Internet references

Math Reference: Trig Substitutions  

Paul's Online Math Notes: Calc II - Trig Substitutions  

Related pages in this website

Table of Integrals


The webmaster and author of this Math Help site is Graeme McRae.