Trig integral strategies Formulas: ∫ Pythagorean identities Double-angle identities sin2 x + cos2 x = 1 1 2 (1 − cos(2x)) = sin2 (x) tan2 x + 1 = sec2 x 1 2 (1 + cos(2x)) = cos2 (x) sina x cosb x dx (1) If either power is odd: Pull out a single power of the/a trig function with the odd power and use the Pythagorean identities to write the remaining powers of that trig function in terms of the other trig function. You are now primed to set u to be that other trig function. Example: ∫ ∫ 3 sin2 x cos8 x sin xdx 8 sin x cos xdx = ∫ (1 − cos2 x) cos8 x sin xdx = ∫ du =∫− sin xdx (1 − cos2 x) cos8 xsin xdx = − (1 − u 2 )u 8 du u = cos x, , (2) If both powers are even: Use the Pythagorean identities to write everything in terms of one trig function. Then repeatedly use the double angle identities until everything is to the ﬁrst power. Example: ∫ ∫ 2 2 sin x cos xdx = ∫ (1 − cos2 x) cos2 xdx (cos2 x − cos4 x)dx ) ∫ ( 1 1 2 = (1 + cos(2x)) − (1 + cos(2x)) dx 2 4 ) ∫ ( 1 1 2 = − cos (2x) dx 4 4 ) ∫ ( 1 1 1 = , − · (1 + cos(4x)) dx 4 4 2 = 1 2 ∫ tana x secb x dx (1) If b is even and > 0: Pull out a sec2 x and use the Pythagorean identities to turn the remaining sec’s into tan’s. You’re now set for the substitution u = tan x , du = sec2 xdx . Example: ∫ ∫ 9 4 tan9 xsec2 x sec2 xdx tan x sec xdx = ∫ = tan9 x(tan2 x + 1) sec2 xdx 2 du = ∫ sec xdx tan9 x(tan2 x + 1)sec2 xdx = u 9 (u 2 + 1)du u = tan x, ∫ , (2) If b = 0: Pull out a tan2 x and switch it using the Pythagorean identity. You’ll get something that falls in case 1 above minus a smaller power of tan. Solve the case 1 ∫piece as above and ∫ apply the same process to the smaller power of tan until you either get 1dx or tan dx , both of which are doable. Example: ∫ ∫ 3 tan2 x tan xdx tan xdx = ∫ (sec2 x − 1) tan xdx ∫ ∫ 2 = tan x sec xdx − tan xdx = , (3) If a is odd: If b = 0, then that’s just case 2 above. For b > 0, pull out a sec x tan x and use the Pythagorean identity to change all the tan’s into sec’s. You are now set to use the substitution u = sec x , du = sec x tan xdx . Example: ∫ ∫ 5 8 tan4 x sec7 x sec x tan xdx tan x sec xdx = ∫ = ∫ u = sec x, (sec2 x − 1)2 sec7 x sec x tan xdx du = sec∫x tan xdx (sec2 x − 1)2 sec7 xsec x tan xdx = (u 2 − 1)2 u 7 du , (4) If a is even and b is odd §: Use the Pythagorean identity to swap all the tan’s into sec’s. You will get a sum of odd powers of sec x . Then use a formula sheet.