FORMULAS TO KNOW Some trig identities: sin2x + cos 2x = 1 tan2x
sin 2x = 2 sin x cos x cos 2x = 2 cos2x − 1 tan x = sin x cos x sec x = 1 cos x cot x = cos x sin x csc x = 1 sin x. Some integration formulas:.
DOUBLE-ANGLE AND HALF-ANGLE IDENTITIES
Third double-angle identity for cosine. Summary of Double-Angles. • Sine: sin 2x = 2 sin x cos x. • Cosine
Example:limit sin^2 x cos^ 2 x dx
1 - cos(2x). 1 + cos(2x). 2. 2. 1 - cos2(2x). = 4. We still have a square so we're still not in the easy case. But this is an.
USEFUL TRIGONOMETRIC IDENTITIES
sin(2x) = 2 sinxcosx cos(2x) = (cosx). 2 - (sinx)2 cos(2x) = 2(cosx). 2 - 1 cos(2x)=1 - 2(sinx). 2. Half angle formulas. [sin(1. 2 x)]. 2. = 1. 2. (1 - cosx). [
TRIGONOMETRY
sin(2x)=2sin(x) cos(x) cos(2x) = cos 2(x) - sin2(x). = 2 cos2(x) 1. = 1 - 2 sin2. -. (x). 2 tan(x) tan(2x) = 1 - tan2(x). HALF-ANGLE IDENTITIES r. ⇣ ⌘ x. 1
MATH 1010 University Mathematics 2014-15.jnt
• √(& Sin²³2x) ( 1 + cos2x) de. 2. = + √ sin² 2x dx + + √ sin² 2x cos 2x dx. case 3 again. =76f1-cas 4x dx + √ √ sin²2x + d sin 2x. -7- Sin 4x+. =7-sinu.
show differential equation 1. If y = x sin 2x prove that x dx2
2018年1月19日 = (2x cos 2x + 2xcos2x − 4x2 sin2x) − (2sin 2x + 4xcos2x) +. 2x sin ... √sin u(− sin u)−cos u cos u. 2√sin u sin u. = 1. 4. −2sin2 u−cos2 ...
1 Trigonometric formula
The cos3(2x) term is a cosine function with an odd power requiring a substitution as done before. We integrate each in turn below. ∫ cos(2x) dx = 1. 2 sin(2x)
Exercise on Integration - Substitution
- sin 2x - cos 2x + C. 13. (sin(ln x) — cos (ln x)) + C. 2. 14. sin 4x1x cos 4x + C. 16. + sin. 1 x x√1-x2. 4. + C. 2x. 5. (2x² + 2x + 1) + C. 15. x2 cos. 2. X.
FORMULAS TO KNOW Some trig identities: sin2x + cos 2x = 1 tan2x
sin 2x = 2 sin x cos x cos 2x = 2 cos2x ? 1 tan x = sin x cos x sec x = 1 cos x cot x = cos x sin x csc x = 1 sin x. Some integration formulas:.
Double-Angle Power-Reducing
https://www.alamo.edu/contentassets/35e1aad11a064ee2ae161ba2ae3b2559/analytic/math2412-double-angle-power-reducing-half-angle-identities.pdf
Techniques of Integration
204 Chapter 10 Techniques of Integration. EXAMPLE 10.1.2 Evaluate ? sin. 6 x dx. Use sin2 x = (1 - cos(2x))/2 to rewrite the function:.
Techniques of Integration
so. ? 2x cos(x2) dx = sin(x2) + C. Even when the chain rule has “produced” a certain derivative it is not always easy to see. Consider this problem:.
Method of Undetermined Coefficients (aka: Method of Educated
yp(x) = A cos(2x) + B sin(2x) where A and B are constants to be determined. Plugging this into the differential equation:.
Example:limit sin^2 x cos^ 2 x dx
Example: sin2 x cos 2 x dx. To integrate sin2 x cos2 x we once again use the half angle formulas: cos 2 ? = 1 + cos(2?). 2 sin2 ? = 1 - cos(2?).
Trigonometric equations
cos x x. Figure 2. A graph of cosx. Example. Suppose we wish to solve sin 2x = ?3. 2for 0 ? x ? 360?. Note that in this case we have the sine of a
USEFUL TRIGONOMETRIC IDENTITIES
sin(2x) = 2 sinxcosx cos(2x) = (cosx). 2 - (sinx)2 cos(2x)=1 - 2(sinx). 2. Half angle formulas. [sin(1. 2 x)]. 2. = 1. 2. (1 - cosx). [cos(1.
Students Solutions Manual
(e) If y = c1 sin 2x + c2 cos 2x then y = 2c1 cos 2x ? 2c2 sin 2x and y = ?4c1 sin 2x ? 4c2 cos 2x = ?4y so y + 4y = 0. (f) If y = c1e2x +c2e?2x
C3 Trigonometry - Trigonometric equations.rtf
sin x + ?3 cos x = 2 sin 2x. (3). (c) Deduce from parts (a) and (b) that sec x + ?3
![Techniques of Integration Techniques of Integration](https://pdfprof.com/Listes/27/34190-27calculus_late_10_Techniques_of_Integration.pdf.pdf.jpg)
Techniques of Integration
Functions consisting of products of the sine and cosine can be integrated by using substi- tution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach.EXAMPLE 10.1.1Evaluate?
sin5xdx. Rewrite the function:
sin5xdx=?
sinxsin4xdx=? sinx(sin2x)2dx=? sinx(1-cos2x)2dx.Now useu= cosx,du=-sinxdx:
sinx(1-cos2x)2dx=? -(1-u2)2du -(1-2u2+u4)du =-u+23u3-15u5+C
=-cosx+23cos3x-15cos5x+C.
203204Chapter 10 Techniques of Integration
EXAMPLE 10.1.2Evaluate?
sin6xdx. Use sin2x= (1-cos(2x))/2 to rewrite the
function: sin6xdx=?
(sin2x)3dx=?(1-cos2x)3
8dx 1 8?1-3cos2x+ 3cos22x-cos32xdx.
Now we have four integrals to evaluate:
1dx=x and -3cos2xdx=-32sin2x
are easy. The cos32xintegral is like the previous example:?
-cos32xdx=? -cos2xcos22xdx -cos2x(1-sin22x)dx -12(1-u2)du
=-1 2? u-u33? =-1 2? sin2x-sin32x3? And finally we use another trigonometric identity, cos2x= (1 + cos(2x))/2:?
3cos22xdx= 3?1 + cos4x
2dx=32?
x+sin4x4?So at long last we get
sin6xdx=x
8-316sin2x-116?
sin2x-sin32x3? +316?x+sin4x4? +C.
EXAMPLE 10.1.3Evaluate?
sin2xcos2xdx. Use the formulas sin2x= (1-cos(2x))/2
and cos2x= (1 + cos(2x))/2 to get:? sin2xcos2xdx=?1-cos(2x)
2·1 + cos(2x)2dx.
The remainder is left as an exercise.
10.2 Trigonometric Substitutions205
Exercises 10.1.
Find the antiderivatives.
1.? sin2xdx?2.?
sin 3xdx? 3. sin4xdx?4.?
cos2xsin3xdx?
5. cos3xdx?6.?
sin2xcos2xdx?
7. cos3xsin2xdx?8.?
sinx(cosx)3/2dx? 9. sec2xcsc2xdx?10.?
tan3xsecxdx?
So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a "reverse" substitution, but it is really no different in principle than ordinary substitution.EXAMPLE 10.2.1Evaluate??
1-x2dx. Letx= sinusodx= cosudu. Then
1-x2dx=?
?1-sin2ucosudu=?⎷cos2ucosudu.We would like to replace
cos2uby cosu, but this is valid only if cosuis positive, since⎷ cos2uis positive. Consider again the substitutionx= sinu. We could just as well think of this asu= arcsinx. If we do, then by the definition of the arcsine,-π/2≤u≤π/2, so cosu≥0. Then we continue: cos2ucosudu=? cos2udu=?1 + cos2u2du=u2+sin2u4+C
arcsinx2+sin(2arcsinx)4+C.
This is a perfectly good answer, though the term sin(2arcsinx) is a bit unpleasant. It isquotesdbs_dbs2.pdfusesText_3[PDF] sites ew
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