sin integral is -cos This is so because the derivative of cos x = -sin x
The integral of cosine cubed is sinx- 1/3 sin cubed x + c
Not correct. sin2alpha + cos2alpha = 1
Cos theta squared
If the angle is measured in degrees, then cos(33) = 0.8387, approx.
7
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
-cos x + Constant
The integral of cos 5x is 1/5 sin (5x)
The integral of x cos(x) dx is cos(x) + x sin(x) + C
sin2x + c
(5.4 / k) cos(kt)
-cos(x) + constant
- cos(1 - X) + C
∫ cos(x) dx = -sin(x) + C
integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C
Ok, I know that sin 2x can be substituted out for 2 sin x cos x. so now I have the Integral of (sin x ) ( 2 sin x cos x ) dx which is 2 sin2x cos x dx If I use integration by parts with the u and dv, I find myself right back again..can you help. The book gives an answer, but I am not sure how it was achieved. the book gives 2/3 sin 2x cos x - 1/3 cos 2x sin x + C I may be making this more difficult than it is ? when you get the integral of 2 sin2x cos x dx use u substitution. u= sinx du= cosxdx. Then you'll get the integral of 2u^2 du.. .and then integrate...
convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx