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Integral Cases for Trigonometric Powers Cheat Sheet by

Algebraic and trigonometric steps to evaluate integrals involving powers of different trigonometric functions. For problems involving powers of trigonometric functions that are both less than 4, other methods may be simpler. For large powers, reduction formulas are recommended.

Lone Cases

Case Name
Step 1 - Exponent Manipu­lation
Step 2 - Trigon­ometric Identities
Step 3
Step 4
Sine-Odd
Manipulate to sin(x)(sin2(x))p
Sine Pythag­orean Id.
Let u=cos(x)
Sine-Even
Manipulate to (sin2(x))p
Sine Power-­Red­ucing Id.
Expand and apply integral linearity
Solve cosine cases
Cosine-Odd
Manipulate to cos(x)(cos2(x))p
Cosine Pythag­orean Id.
Let u=sin(x)
Cosine­-Even
Manipulate to (cos2(x))p
Cosine Power-­Red­ucing Id.
Expand and apply integral linearity
Solve cosine cases
Tangen­t-Odd
Manipulate to tan(x)(tan2(x))p
Tangent Pythag­orean Id.
Expand and apply integral linearity
Solve other cases
Tangen­t-Even
Manipulate to (tan2(x))p
Tangent Pythag­orean Id.
Expand and apply integral linearity
Solve secant cases
Coseca­nt-Odd
Manipulate to csc(x)(csc2(x))p
Cosecant Pythag­orean Id.
Expand and apply integral linearity
Solve other cases
Coseca­nt-Even
Manipulate to csc2(x)(csc2(x))p
Cosecant Pythag­orean Id.
Let u=cot(x)
Secant-Odd
Manipulate to sec(x)(sec2(x))p
Secant Pythag­orean Id.
Expand and apply integral linearity
Solve other cases
Secant­-Even
Manipulate to sec2(x)(sec2(x))p
Secant Pythag­orean Id.
Let u=tan(x)
Cotang­ent-Odd
Manipulate to cot(x)(cot2(x))p
Cotangent Pythag­orean Id.
Expand and apply integral linearity
Solve other cases
Cotang­ent­-Even
Manipulate to (cot2(x))p
Cotangent Pythag­orean Id.
Expand and apply integral linearity
Solve cosecant cases
p ∈ ℕ1 = {1,2,3­­­,­4­,­­5­­,...}

Pair Cases

Case Name
Step 1 - Exponent Manipu­lation
Step 2
Step 3
Step 4
Sine-Odd and Cosine­-Even
Manipulate sine powers to sin(x)(sin2(x))p
Sine Pythag­orean Id.
Let u=cos(x)
Sine-Even and Cosine-Odd
Manipulate sine powers to cos(x)(cos2(x))p
Cosine Pythag­orean Id.
Let u=sin(x)
Sine-Odd and Cosine-Odd
Either two previous cases work
Sine-Even and Cosine­-Even
Manipulate sine powers to (sin2(x))p
Sine Pythag­orean Id.
Expand and apply integral linearity
Solve cosine cases
Sine-C­osine Same
Manipulate powers to (sin(x­)co­s(x))p
Sine Double­-Angle Id.
Solve sine cases
Secant-Odd and Tangen­t-Even
Manipulate tangent powers to (tan2(x))p
Tangent Pythag­orean Id.
Expand and apply integral linearity
Solve secant cases
Secant­-Even and Tangen­t-Odd
Manipulate powers to sec(x)­tan­(x)secp(x)tanq(x)
Manipulate tangent powers to (tan2(x))r
Tangent Pythag­orean Id.
Let u=sec(x)
Secant-Odd and Tangen­t-Odd
Same as previous case
Secant­-Even and Tangen­t-Even
Manipulate secant powers to sec2(x)(sec2(x))p
Secant Pythag­orean Id.
Let u=tan(x)
Secant­-Ta­ngent Same
Same as previous two cases
Coseca­nt-Odd and Cotang­ent­-Even
Manipulate cotangent powers to (cot2(x))p
Cotangent Pythag­orean Id.
Expand and apply integral linearity
Solve cosecant cases
Coseca­nt-Even and Cotang­ent-Odd
Manipulate powers to csc(x)­cot­(x)cscp(x)cotq(x)
Manipulate cotangent powers to (cot2(x))r
Cotangent Pythag­orean Id.
Let u=csc(x)
Coseca­nt-Odd and Cotang­ent-Odd
Same as previous case
Coseca­nt-Even and Cotang­ent­-Even
Manipulate cosecant powers to csc2(x)(csc2(x))p
Cosecant Pythag­orean Id.
Let u=cot(x)
Coseca­nt-­Cot­angent Same
Same as previous two cases
p, q, r ∈ ℕ1 = {1,2,3­­­,­4­,­­5­­,...}

Pythag­orean Identities

Sine Pythag­orean
sin2(x)=1-cos2(x)
Cosine Pythag­orean
cos2(x)=1-sin2(x)
Tangent Pythag­orean
tan2(x)=sec2(x)-1
Cosecant Pythag­orean
csc2(x)=cot2(x)+1
Secant Pythag­orean
sec2(x)=tan2(x)+1
Cotangent Pythag­orean
cot2(x)=csc2(x)-1
Sine-C­osine Pythag­orean
1=sin2(x)+cos2(x)
Secant­-Ta­ngent Pythag­orean
1=sec2(x)-tan2(x)
Coseca­nt-­Cot­angent Pythag­orean
1=csc2(x)-cot2(x)

Reduction Formulas

∫sinn(x)dx, n≥2
-cos(x)sinn-1(x)/n + (n-1/n­)∫sinn-2(x)dx
∫cosn(x)dx, n≥2
cos(x)n-1sin(x)/n + (n-1/n­)∫cosn-2(x)dx
∫tann(x)dx, n≥2
tann-1(x)/(n-1) - ∫tann-2(x)dx
∫cscn(x)dx, n≥2
-cscn-2(x)cot­(x)­/(n-1) + ((n-2)­/(n­-1)­)∫cscn-2(x)dx
∫secn(x)dx, n≥2
secn-2(x)tan­(x)­/(n-1) + ((n-2)­/(n­-1)­)∫secn-2(x)dx
∫cotn(x)dx, n≥2
-cotn-1(x)/(n-1) - ∫cotn-2(x)dx
n ∈ ℕ1 = {1,2,3­­­,­4­,­­5­­,...}
  

Miscel­laneous Inform­ation

Binomial Expansion
(a+b)n = ∑nk=0 (nk) an-kbk
Integral Linearity Property: Sum and Difference
∫(f(x) ± g(x))dx = ∫f(x)dx ± ∫g(x)dx
∫tan(x)dx
ln|sec­(x)|+C
∫cot(x)dx
ln|sin­(x)|+C
∫sec(x)dx
ln|sec­(x)­+ta­n(x)|+C
∫sec3(x)dx
½(sec(­­x)­­­t­an­­­(x­­)­­+­l­n­­­|se­­­c­(­x­­)­­+t­­­an­(­x­)|)+C
∫csc(x)dx
ln|csc­(x)­-co­t(x)|+C
∫csc3(x)dx
-½(csc­(­x­)­­­cot­­­(­x­)­­+­­ln­­­|c­s­c­(­x­­)­­+co­t(­­x)|)+C
(nk) are the binomial coeffi­cients, equal to n!/(k!­(n-k)!)
n, k ∈ ℕ1 = {1,2,3­­­,­4­,­­5­­,...}
                           
 

Metadata

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  • Published: 6th July, 2024
  • Last Updated: 7th October, 2024

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