[PDF] Techniques of Integration





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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 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 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 10

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?

sin

5xdx. Rewrite the function:

sin

5xdx=?

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+2

3u3-15u5+C

=-cosx+2

3cos3x-15cos5x+C.

203

204Chapter 10 Techniques of Integration

EXAMPLE 10.1.2Evaluate?

sin

6xdx. Use sin2x= (1-cos(2x))/2 to rewrite the

function: sin

6xdx=?

(sin

2x)3dx=?(1-cos2x)3

8dx 1 8?

1-3cos2x+ 3cos22x-cos32xdx.

Now we have four integrals to evaluate:

1dx=x and -3cos2xdx=-3

2sin2x

are easy. The cos

32xintegral is like the previous example:?

-cos32xdx=? -cos2xcos22xdx -cos2x(1-sin22x)dx -1

2(1-u2)du

=-1 2? u-u33? =-1 2? sin2x-sin32x3? And finally we use another trigonometric identity, cos

2x= (1 + cos(2x))/2:?

3cos

22xdx= 3?1 + cos4x

2dx=32?

x+sin4x4?

So at long last we get

sin

6xdx=x

8-316sin2x-116?

sin2x-sin32x3? +316?
x+sin4x4? +C.

EXAMPLE 10.1.3Evaluate?

sin

2xcos2xdx. Use the formulas sin2x= (1-cos(2x))/2

and cos2x= (1 + cos(2x))/2 to get:? sin

2xcos2xdx=?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.? sin

2xdx?2.?

sin 3xdx? 3. sin

4xdx?4.?

cos

2xsin3xdx?

5. cos

3xdx?6.?

sin

2xcos2xdx?

7. cos

3xsin2xdx?8.?

sinx(cosx)3/2dx? 9. sec

2xcsc2xdx?10.?

tan

3xsecxdx?

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=? cos

2udu=?1 + cos2u2du=u2+sin2u4+C

arcsinx

2+sin(2arcsinx)4+C.

This is a perfectly good answer, though the term sin(2arcsinx) is a bit unpleasant. It isquotesdbs_dbs2.pdfusesText_3
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