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:

and, because sec²θ - 1 = tan²θ,

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

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:

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

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