# 1 5 1 7 u 7 c 1 25 sin 5 5 x 1 35 sin 7 5 x c 620

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Chapter 13 / Exercise 77
Multivariable Calculus
Larson
Expert Verified
15·17u7+C=125sin55x135sin75x+C.
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Chapter 13 / Exercise 77
Multivariable Calculus
Larson
Expert Verified
620CHAPTER 7.ADVANCED INTEGRATION TECHNIQUESThe technique works even if only one of the trigonometric functions sine or cosine appears,as long as it is to an odd power.Example 7.3.6Computeintegraldisplaysin37x dx.Solution: Here we can still peel off a sine factor to be the functional part of our differential,and then write the remaining factors in terms of the cosine.integraldisplaysin37x dx=integraldisplaysin27xsin 7x dx=integraldisplay(1cos27x)sin 7x dx.Using the substitutionu= cos 7x, sodu=7 sin 7x dx, implying17du= sin 7x dx, we getintegraldisplaysin37x dx=integraldisplay(1cos27x)sin 7x dx=integraldisplay(1u2)·17du=17bracketleftbiggu13u3bracketrightbigg+C=17cos 7x+121cos37x+C.Furthermore, not all the powers need to be positive integer powers, as long as one is odd.Example 7.3.7Computeintegraldisplaycos7xsinxdx.Solution: Here we have an odd number of cosine terms, so we will peel one off to be thefunctional part of our differential. That is, we will haveu= sinx, sodu= cosx dx. Thusintegraldisplaycos7xsinxdx=integraldisplaycos6xsinxcosx dx=integraldisplay(1sin2x)3sinxcosx dx=integraldisplay(1u2)3udu=integraldisplay13u2+ 3u4u6u1/2du=integraldisplaybracketleftBigu1/23u3/2+ 3u7/2u11/2bracketrightBigdu= 2u1/23·25u5/2+ 3·29u9/2213u13/2+C= 2u1/2bracketleftbigg135u2+13u4113u6bracketrightbigg+C= 2sinxbracketleftbigg135sin2x+13sin4x113sin6xbracketrightbigg+C.
7.3.TRIGONOMETRIC INTEGRALS621It is possible that both powers are odd, and either function can be peeled off, and the integralwritten in terms of the other. However, if one of these odd powers is greater than the other, itis more efficient to peel off a factor from the lower power, as the next example demonstrates.Example 7.3.8Computeintegraldisplaysin3xcos7x dx.Solution: We will consider both methods for computing this antiderivative. First we peel offa sine to be part of the differential, and letu= cosx(sodu=sinx dx).integraldisplaysin3xcos7x dx=integraldisplaysin2xcos7xsinx dx=integraldisplay(1cos2x)cos7xsinx dx=integraldisplay(1u2)u7(du)=integraldisplay(u7u9)du=18u8+110u10+C=18cos8x+110cos10x+C.Next we instead peel off a cosine factor, and letw= sinx(so thatdw= cosx dx).integraldisplaysin3xcos7x dx=integraldisplaysin3xcos6xcosx dx=integraldisplaysin3x(1sin2x)3cosx dx=integraldisplayw3(1w2)3dw=integraldisplayw3(13w2+ 3w4w6)dw=integraldisplay(w33w5+ 3w7w9)dw=14w436w6+38w8110w10+C=14sin4x12sin6x+38sin8x110sin10x+C.As we see in the above example, there can be different valid choices for some integrals. The