Cheatography

# Integral Cases for Trigonometric Powers Cheat Sheet by CROSSANT

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 Manipulate to sin(x)(sin2(x))p Sine Pythag­orean Id. Let u=cos(x) Sine-Even and Cosine-Odd Manipulate sine powers to Manipulate 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,...}