Math 141Exam Questions with100% Correct
Answers
∫sin(x) dx
-cos(x) + c
∫cos(x) dx
sin(x) + c
∫sec^2 (x) dx
tan(x) + C
∫ sec(x)tan(x) dx
sec(x)+ c
∫ csc^2(x) dx
-cotx + c
∫cscxcotx dx
-csc(x) + C
∫ tan(x) dx
ln|sec(x)| + c
∫ sec(x) dx
ln|sec(x)+tan(x)| + c
∫ csc(x) dx
, ln|cscx-cotx|+c or -ln|cscx+cotx|
∫ 1/(x^2 + 1) dx
arctanx + C
∫ 1/sqrt(1-x^2) dx
arcsinx + C
∫ -1/sqrt(1-x^2) dx
arccosx + C
pythagorean identies
sin^2 (x) + cos^2 (x) = 1
sin(x)cos(y)
1/2[sin(x-y)+sin(x+y)]
sin(x)sin(y)
1/2[cos(x-y)-cos(x+y)]
cos(x)cos(y)
1/2[cos(x-y)+cos(x+y)]
sin^2 (x)
(1 - cos 2x) / 2
cos^2 (x)
Answers
∫sin(x) dx
-cos(x) + c
∫cos(x) dx
sin(x) + c
∫sec^2 (x) dx
tan(x) + C
∫ sec(x)tan(x) dx
sec(x)+ c
∫ csc^2(x) dx
-cotx + c
∫cscxcotx dx
-csc(x) + C
∫ tan(x) dx
ln|sec(x)| + c
∫ sec(x) dx
ln|sec(x)+tan(x)| + c
∫ csc(x) dx
, ln|cscx-cotx|+c or -ln|cscx+cotx|
∫ 1/(x^2 + 1) dx
arctanx + C
∫ 1/sqrt(1-x^2) dx
arcsinx + C
∫ -1/sqrt(1-x^2) dx
arccosx + C
pythagorean identies
sin^2 (x) + cos^2 (x) = 1
sin(x)cos(y)
1/2[sin(x-y)+sin(x+y)]
sin(x)sin(y)
1/2[cos(x-y)-cos(x+y)]
cos(x)cos(y)
1/2[cos(x-y)+cos(x+y)]
sin^2 (x)
(1 - cos 2x) / 2
cos^2 (x)
