Calculus 2
sin^2(x)
(1-cox2x)/2
int sinx
-cosx+c
int csc^2(x)
-cotx+c
d/dx cotx
-csc^2(x)
int cscxcotx
-cscx + C
d/dx cscx
-cscxcotx
d/dx cosx
-sinx
cos^2(x)
1/2 + 1/2cos2x
sin^2(x)
1/2 - 1/2cos2x
int 1/(u)sqrt(u^2-a^2)
1/a[arcsec(u/a)] + C
Int 1/a^2+u^2
1/a[arctan(u/a)] +C
d/dx ln(u)
1/u • du/dx
Arc Length
L= int: sqrt ( (dy/dt)^2 + (dx/dt)^2) dy
Washer Method
V= int: piRouter^2 - piRinner^2 dx
Disk Method
V= int: pi[f(x)]^2 dx
LHopital
When lim fx and lim gx equals zero or pos/neg infinity. Lim of fx/gx equals lim f'x/g'x.
int 1/sqrt(a^2-u^2)
arcsin(u/a)+C
cos(2x)
cos^2(x) - sin^2(x)
d/dx sinx
cosx
1 + cot^2(x)
csc^2(x)
integrals of partial fractions will always work when
denominator factors into linear functions and degree of numerator is lower than degree of denominator --> if not do long division and test the Remainder
d/dx e^(f(x))
e^(f(x)) • f'(x)
d/dx e^x
e^x
int e^x dx
e^x + c
if a sereis is bounded and monotonic
it will always converge
series converges if
limit exists as a finite number
int (1/u) du
ln |u| + c
int cscx
ln|cscx-cotx| + C
int secx
ln|secx+tanx|+C
int tanx
ln|secx|+C
int cotx
ln|sinx| + C
d/dx tanx
sec^2(x)
tan^2(x) + 1
sec^2(x)
tan^2(x)+1
sec^2(x)
int secxtanx
secx+c
d/dx secx
secxtanx
if lim of an doesn't go to zero
series cannot converge
int cosx
sinx+c
int sec^2(x)
tanx + C
Integration by parts
uv - int: vdu
Trig Sub: sqrt(x^2 - a^2)
x = asec(x) and sec^2 - 1 = tan^2
Trig Sub: sqrt(a^2 - x^2)
x = asin(x) and 1 - sin^2 = cos^2
Trig Sub: sqrt( a^2+x^2)
x = atan(x) and 1 + tan^2(x) = sec^2