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Integration Using Trig Substitution ~ p. 1
INTEGRATION USING TRIG SUBSTITUTION
Objective: Integrate Using Trig Substitution
Pythagorean trig identity : sin^2 x^ +^ cos^2 x=^1
From this identity we can quickly derive the other two related identities
sin sin
cos sin sin
2 2
2 2 2
x 1 x
x x x
cos cos cos
2 2
2 2 2
x 1 x
x x x
Example: Evaluate (^) ∫cos^3 xdx
- Neither substitution nor integration by parts will work here because a factor of (sin x) would be introduced.
- Separate the cosine factor into a product and substitute using the Pythagorean identity
- (^) ∫ cos^3 xdx = (^) ∫ (cos^2 x • cos x dx) =∫ ( 1 − sin^2 x) • cos x dx)
∫ (^1 −^ sin^2 x)^ • cos^ x dx)^ =^ ∫ (^1 −u^2 )du^ let u = sin x ⇒^ du^ =cosx
∫ (^1 −u^2 )du^ =^ u^ −^ u^ +C =
3 3
sin x − sin x^ +C
3 3
- General rule: Try to rewrite integrand involving odd powers of sine and/or cosine in a form which has a single sine factor or a single cosine factor. [see next section if the integrand has only even powers of sine and/or cosine.
Evaluate (^) ∫sin 3 xdx
If integrand contains only even powers of sine and/or cosine, use the following half-angle
identities
- sin^2 x = 12 ( 1 −cos 2 x)
- cos^2 x = 21 ( 1 +cos 2 x)
Integration Using Trig Substitution ~ p. 2
Example: Evaluate (^) ∫cos^4 xdx
- (^) ∫ cos^4 xdx= (^) ∫ (cos^2 x) 2 dx = (^) [ 12 ] =
2 ∫ (^1 +cos^2 x)^ dx^41 ∫ (^1 +^2 cos^2 x^ +cos^2 2 x dx) = 41 x + 41 sin 2 x + (^41) ∫ 21 ( 1 +cos 4 x dx) = 41 x + 41 sin 2 x + 81 x + (^81) ( 41 sin 4 x) +C
= 38 x + 41 sin 2 x + 321 sin 4 x +C
Evaluate (^) ∫sin 4 x dx
A trig substitution may sometimes be used to get rid of a root sign: see Example 3 on p. 401
- If you have a factor of a^2 −^ x^2 , let x = a(sin θ)
- If you have a factor of a^2 +^ x^2 , let x =a(tan θ)
- If you have a factor of x^2 −^ a^2 , let x =a(sec θ)
- Example: Evaluate if
2 0
4 −x 2 dx
∫ (^02)
≤ θ≤^ π
- Let x = 2(sin θ ), then dx =2(cos θ )dθ
and a 2 − x 2 = 2 2 − 2 2 sin 2 θ= 2 2 (1 − sin 2 θ) = 2 2 cos^2 θ
- So, (^) ∫ 2 2 − x 2 dx =∫ ( 2 cos θ)^2 2cos θ dθ = ∫ (2 cos θ)(2 cos θ )d θ =∫4 cos 2 θ dθ
- We now have (^4) ∫cos 2 θ dθ, which we can integrate using a half angle formula.
- (^4) ∫ cos 2 θ dθ = (^4) ∫^12 (1 + cos 2 θ ) d θ = 2( θ+^12 sin 2 θ)
- It is easier to convert the limits than to change everything back in terms of x!
- If x = 0, then θ = sin −^1 2 x= sin −^10 = 0
- If x = 2, then 1 2 1 2
θ = sin −^ x= sin − 1 =^ π
- 2( 21 sin 2 ) (^022) ( 2 12 sin (^) ) 2 0( 12 sin 0)
π (^) π
If you are finding an indefinite integral , you must convert back to the original variable! Problems
Integration Using Trig Substitution ~ p. 4
Solutions
• ∫ sin 3 xdx = ∫(1 −cos 2 x)sin x dx Let u = cos x; then du = –sinx dx
= (1 u 2 )du ( u 1 u 3 ) C
= cos x 1 cos 3 x C or 1 cos 3 x cos x C
• ∫ sin 4 x dx= ∫ (sin 2 x) 2 dx = =
1 2
∫ ⎡⎣^2 (1^ −cos 2x)^ ⎤⎦ dx
1 2
4 ∫(1^ −^ 2cos 2x^ +cos^ 2x)dx
= 41 x − 14 sin 2x + 14 ∫ 21 (1 +cos 4x)dx =^14 x − 14 sin 2x + 18 x + 18 ( 14 sin 4x )+C
= 38 x − 14 sin 2x + 321 sin 4x +C
