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Searches related to arcsinsinx filetype:pdf 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 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 is possible to simplify this. Using the identity sin2x= 2sinxcosx, we can write sin2u=

2sinucosu= 2sin(arcsinx)?

1-sin2u= 2x?1-sin2(arcsinx) = 2x?1-x2.Then the

full antiderivative is arcsinx

2+2x⎷

1-x2

4=arcsinx2+x⎷

1-x2 2+C.

206Chapter 10 Techniques of Integration

This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to usethe fundamental identity sin

2x+ cos2x= 1 in one of three forms:

cos

2x= 1-sin2xsec2x= 1 + tan2xtan2x= sec2x-1.

If your function contains 1-x2, as in the example above, tryx= sinu; if it contains 1+x2 tryx= tanu; and if it containsx2-1, tryx= secu. Sometimes you will need to try something a bit different to handle constants other than one.

EXAMPLE 10.2.2Evaluate??

4-9x2dx. We start by rewriting this so that it looks

more like the previous example:??

4-9x2dx=??4(1-(3x/2)2)dx=?

2?1-(3x/2)2dx.

Now let 3x/2 = sinuso (3/2)dx= cosuduordx= (2/3)cosudu. Then? 2?

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

2?1-sin2u(2/3)cosudu=43?

cos 2udu 4u

6+4sin2u12+C

2arcsin(3x/2)

3+2sinucosu3+C

2arcsin(3x/2)

3+2sin(arcsin(3x/2))cos(arcsin(3x/2))3+C

2arcsin(3x/2)

3+2(3x/2)?

1-(3x/2)2

3+C

2arcsin(3x/2)

3+x⎷

4-9x2 2+C, using some of the work from example 10.2.1. EXAMPLE 10.2.3Evaluate??1 +x2dx. Letx= tanu,dx= sec2udu, so

1 +x2dx=?

?1 + tan2usec2udu=?⎷sec2usec2udu. sec2u= secu. Then?⎷ sec2usec2udu=? sec 3udu. In problems of this type, two integrals come up frequently: sec

3uduand?

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