# Trig integral strategies ```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
=
,
− &middot; (1 + cos(4x)) dx
4 4 2
=
1
2
∫
tana x secb x dx
(1) If b is even and &gt; 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 &gt; 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 &sect;: 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.
```