Section 7.3: Trigonometric Substitution

Review

To integrate ∫(sin^n)(x)(cos^m)(x)d x

Integrate[Sin[x]^n Cos[x]^m,x]

If n is a positive odd integer, write this integral as
	∫(sin^n)(x)(cos^m)(x)d x=∫(sin^(n-1))(x)(cos^m)(x)sin(x)d x
and use the substitution u=cos(x) so that d u=-sin(x)d x. Since n is odd, n-1 is even so that n-1=2k.
	(sin^(n-1))(x)=(sin^(2k))(x)=((((sin^2)(x)))^k)=(((1-(cos^2)(x)))^k).

If m is a positive odd integer, do the same thing as the above using the substitution u=sin(x).

Question

To integrate ∫(sec^n)(x)(tan^m)(x)d x

Integrate[Sec[x]^n Tan[x]^m,x]

If n is a positive even integer, write this integral as
	∫(sec^n)(x)(tan^m)(x)d x=∫(sec^(n-2))(x)(tan^m)(x)(sec^2)(x)d x
and use the substitution u=tan(x) so that d u=(sec^2)(x)d x. Since n is even, n-2 is even so that n-2=2k.
	(sec^(n-2))(x)=(sec^(2k))(x)=((((sec^2)(x)))^k)=(((1+(tan^2)(x)))^k).

If m is a positive odd integer, write this integral as
	∫(sec^n)(x)(tan^m)(x)d x=∫(sec^(n-1))(x)(tan^(m-1))(x)tan(x)sec(x)d x
and use the substitution u=sec(x) so that d u=sec(x)tan(x)d x. Also, use the trigonometric identity (tan^2)(x)=(sec^2)(x)-1.

Question

Evaluate ∫(tan^5)(x)(sec^4)(x)d x.

Integrate[Sec[x]^4 Tan[x]^5,x]

Back to ∫(cos^n)(x)(sin^m)(x)d x:

If  n and m are positive even numbers use the trigonometric identities
	(sin^2)(θ)=((1-cos(2θ))/2)
	(cos^2)(θ)=((1+cos(2θ))/2)
to reduce the power on the trig functions.

Question

What is the value of ∫(sin^2)(3x)d x?

Integrate[Sin[3x]^2,x]

What is the value of ∫(sin^2)(x)(cos^2)(x)d x?

Integrate[Sin[x]^2 Cos[x]^2,x]

Simple trigonometric integrals

You know the values of ∫cos(x)d x and ∫sin(x)d x.

Write tan(x) as ((sin(x))/(cos(x))) and use a substitution to evaluate ∫tan(x)d x.

Integrate[Tan[x],x]

What is ∫cot(x)d x?

Integrate[Cot[x],x]

Write sec(x) as sec(x) ((sec(x)+tan(x))/(sec(x)+tan(x))) and use a substitution to evaluate ∫sec(x)d x.

Integrate[Sec[x],x]

What is ∫csc(x)d x?

Integrate[Csc[x],x]

Trigonometric identities

Pythagorean identities

(cos^2)(θ)+(sin^2)(θ)=1	⟶	(cos^2)(θ)=1-(sin^2)(θ)
This means for any number a>0,
	(Sqrt[(a^2)-(a^2)(sin^2)(θ)])=(Sqrt[(a^2)(cos^2)(θ)])=±a cos(θ)

Questions

Simplify (Sqrt[(a^2)+(a^2)(tan^2)(θ)]).

Simplify (Sqrt[(a^2)(sec^2)(θ)-(a^2)]).

Trigonometric substitutions

If an integral involves (Sqrt[(a^2)-(x^2)]), use the substitution x=a sin(θ).

If an integral involves (Sqrt[(a^2)+(x^2)]), use the substitution x=a tan(θ).

If an integral involves (Sqrt[(x^2)-(a^2)]), use the substitution x=a sec(θ).

Example

To evaluate ∫(x^3)(Sqrt[4-(x^2)])d x use the substitution x=2sin(θ) with -π/2≤θ≤π/2. That means d x=2cos(θ)d θ.
	∫(x^3)(Sqrt[4-(x^2)])d x=∫8(sin^3)(θ)(Sqrt[4-4(sin^2)(θ)])2cos(θ)d θ
	=∫8(sin^3)(θ)(Sqrt[4(cos^2)(θ)])2cos(θ)d θ
	=∫8(sin^3)(θ)2cos(θ)2cos(θ)d θ
	=32∫(sin^3)(θ)(cos^2)(θ)d θ
	=32∫(sin^2)(θ)(cos^2)(θ)sin(θ)d θ
	=32∫(1-(cos^2)(θ))(cos^2)(θ)sin(θ)d θ
Now make the substitution u=cos(θ) so that d u=-sin(θ)d θ or sin(θ)d θ=-d u.
		=-32∫(1-(u^2))(u^2)d u
=-32∫((u^2)-(u^4))d u
=-32(((u^3)/3)-((u^5)/5))+C
=-32((((cos^3)(θ))/3)-(((cos^5)(θ))/5))+C
Since x=2sin(θ), sin(θ)=x/2

Graphics[{Line[{{0,0},{2,0},{2,1},{0,0}}],Text[θ,.4{Cos[.5ArcTan[2,1]],Sin[.5ArcTan[2,1]]}],Text[x,{2.1,.5}],Text[2,{1,.6}]}]

From the corresponding right triangle, cos(θ)=((Sqrt[4-(x^2)])/2). The final answer is
	-32((((cos^3)(θ))/3)-(((cos^5)(θ))/5))+C=-32((((((Sqrt[4-(x^2)])/2))^3)/3)-(((((Sqrt[4-(x^2)])/2))^5)/5))+C
	=-(4/3)((((Sqrt[4-(x^2)])))^3)+(1/5)((((Sqrt[4-(x^2)])))^5)+C

Question

Consider ∫(1/(Sqrt[9+(x^2)]))d x.

Is this a sin(θ), tan(θ), or sec(θ) substitution?

What is the appropriate trig substitution?

After the substitution what is the resulting trigonometric integral?

What is the value of the original integral?

Integrate[1/Sqrt[9+x^2],x]

Question

Use an appropriate trigonometric substitution to find the area between the x-axis, y=((x^3)/(Sqrt[(x^2)-1])) and x=2 and x=3.

Question

Consider ∫(Sqrt[9-4(x^2)])d x.

Is this a sin(θ), tan(θ), or sec(θ) substitution?

What is the appropriate trig substitution?

After the substitution what is the resulting trigonometric integral?

What is the value of the original integral?

Integrate[Sqrt[9-4x^2],x]