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Integration of trigonometric functions problems and solutions pdf

3.2.1 Solve integration problems involving products and powers of sinxsinx and cosx.cosx. 3.2.2 Solve integration problems involving products and powers of tanxtanx and secx.secx. 3.2.3 Use reduction formulas to solve trigonometric integrals. In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals. They are an important part of the integration technique called trigonometric substitution, which is featured in Trigonometric Substitution. This technique allows us to convert algebraic expressions that we may not be able to integrate into expressions involving trigonometric functions, which we may be able to integrate using the techniques described in this section. In addition, these types of integrals appear frequently when we study polar, cylindrical, and spherical coordinate systems later. Let's begin our study with products of sinxsinx and cosx.cosx. A key idea behind the strategy used to integrate combinations of products and powers of sinxsinx and cosxcosx involves rewriting these expressions as sums and differences of integrals of the form sinjxcosxdxsinjxcosxdx or cosjxsinxdx.cosjxsinxdx. After rewriting these integrals, we evaluate them using u-substitution. Before describing the general process in detail, let's take a look at the following examples. Evaluate cos3xsinxdx.cos3xsinxdx. Use uu-substitution and let u=cosx.u=cosx. In this case, du=-sinxdx.du=-sinxdx. Thus, cos3xsinxdx=-u3du=-14u4+C=-14cos4x+C.cos3xsinxdx=-u3du=-14u4+C=-14cos4x+C. Evaluate sin4xcosxdx.sin4xcosxdx. Evaluate cos2xsin3xdx.cos2xsin3xdx. To convert this integral to integrals of the form cosjxsinxdx,cosjxsinxdx, rewrite sin3x=sin2xsinxsin3x=sin2xsinx and make the substitution sin2x=1-cos2x.sin2x=1-cos2x. Thus, cos2xsin3xdx=cos2x(1-cos2x)sinxdxLetu=cosx;thendu=-sinxdx.=-u2(1-u2)du=(u4-u2)du=15u5-13u3+C=15cos5x-13cos3x+C.cos2xsin3xdx=cos2x(1-cos2x)sinxdxLetu=cosx;thendu=-sinxdx.=-u2(1-u2)du=(u4-u2)du=15u5-13u3+C=15cos5x-13cos3x+C. Evaluate cos3xsin2xdx.cos3xsin2xdx. In the next example, we see the strategy that must be applied when there are only even powers of sinxsinx and cosx.cosx. For integrals of this type, the identities sin2x=12-12cos(2x)=1-cos(2x)2sin2x=12-12cos(2x)=1-cos(2x)2 and cos2x=12+12cos(2x)=1+cos(2x)2cos2x=12+12cos(2x)=1+cos(2x)2 are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity cos(2x)=cos2x-sin2xcos(2x)=cos2x-sin2x and the Pythagorean identity cos2x+sin2x=1.cos2x+sin2x=1. Evaluate sin2xdx.sin2xdx. To evaluate this integral, let's use the trigonometric identity sin2x=12-12cos(2x).sin2x=12-12cos(2x). Thus, sin2xdx=(12-12cos(2x))dx=12x-14sin(2x)+C.sin2xdx=(12-12cos(2x))dx=12x-14sin(2x)+C. Evaluate cos2xdx.cos2xdx. The general process for integrating products of powers of sinxsinx and cosxcosx is summarized in the following set of guidelines. To integrate cosjxsinkxdxcosjxsinkxdx use the following strategies: If kk is odd, rewrite sinkx=sink-1xsinxsinkx=sink-1xsinx and use the identity sin2x=1-cos2xsin2x=1-cos2x to rewrite sink-1xsink-1x in terms of cosx.cosx. Integrate using the substitution u=cosx.u=cosx. This substitution makes du=-sinxdx.du=-sinxdx. If jj is odd, rewrite cosjx=cosj-1xcosxcosjx=cosj-1xcosx and use the identity cos2x=1-sin2xcos2x=1-sin2x to rewrite cosj-1xcosj-1x in terms of sinx.sinx. Integrate using the substitution u=sinx.u=sinx. This substitution makes du=cosxdx.du=cosxdx. (Note: If both jj and kk are odd, either strategy 1 or strategy 2 may be used.) If both jj and kk are even, use sin2x=(1/2)-(1/2)cos(2x)sin2x=(1/2)-(1/2)cos(2x) and cos2x=(1/2)+(1/2)cos(2x).cos2x=(1/2)+(1/2)cos(2x). After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate. Evaluate cos8xsin5xdx.cos8xsin5xdx. Since the power on sinxsinx is odd, use strategy 1. Thus, cos8xsin5xdx=cos8xsin4xsinxdxBreak offsinx.=cos8x(sin2x)2sinxdxRewritesin4x= (sin2x)2.=cos8x(1-cos2x)2sinxdxSubstitutesin2x=1-cos2x.=u8(1-u2)2(-du)Letu=cosxanddu=-sinxdx.=(-u8+2u10-u12)duExpand.=-19u9+211u11-113u13+CEvaluate the integral.=-19cos9x+211cos11x-113cos13x+C.Substituteu=cosx.cos8xsin5xdx=cos8xsin4xsinxdxBreak offsinx.=cos8x(sin2x)2sinxdxRewritesin4x= (sin2x)2.=cos8x(1-cos2x)2sinxdxSubstitutesin2x=1-cos2x.=u8(1-u2)2(-du)Letu=cosxanddu=-sinxdx.=(-u8+2u10-u12)duExpand.=-19u9+211u11-113u13+CEvaluate the integral.=-19cos9x+211cos11x-113cos13x+C.Substituteu=cosx. Evaluate sin4xdx.sin4xdx. Since the power on sinxsinx is even (k=4)(k=4) and the power on cosxcosx is even (j=0), (j=0), we must use strategy 3. Thus, sin4xdx=(sin2x)2dxRewritesin4x=(sin2x)2.=(12-12cos(2x))2dxSubstitutesin2x=12-12cos(2x).=(14-12cos(2x)+14cos2(2x))dxExpand(12-12cos(2x))2.=(14-12cos(2x)+14(12+12cos(4x))dx.sin4xdx=(sin2x)2dxRewritesin4x= (sin2x)2.=(12-12cos(2x))2dxSubstitutesin2x=12-12cos(2x).=(14-12cos(2x)+14cos2(2x))dxExpand(12-12cos(2x))2.=(14-12cos(2x)+14(12+12cos(4x))dx. Since cos2(2x)cos2(2x) has an even power, substitute cos2(2x)=12+12cos(4x):cos2(2x)=12+12cos(4x): =(38-12cos(2x)+18cos(4x))dxSimplify.=38x-14sin(2x)+132sin(4x)+CEvaluate the integral.=(38-12cos(2x)+18cos(4x))dxSimplify.=38x-14sin(2x)+132sin(4x)+CEvaluate the integral. Evaluate cos3xdx.cos3xdx. Evaluate cos2(3x)dx.cos2(3x)dx. In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include sin(ax),sin(ax), sin(bx),sin(bx), cos(ax),cos(ax), and cos(bx).cos(bx). These integrals are evaluated by applying trigonometric identities, as outlined in the following rule. To integrate products involving sin(ax),sin(ax), sin(bx),sin(bx), cos(ax),cos(ax), and cos(bx),cos(bx), use the substitutions sin(ax)sin(bx)=12cos((a-b)x)-12cos((a+b)x)sin(ax)sin(bx)=12cos((a-b)x)-12cos((a+b)x) sin(ax)cos(bx)=12sin((a-b)x)+12sin((a+b)x)sin(ax)cos(bx)=12sin((a-b)x)+12sin((a+b)x) cos(ax)cos(bx)=12cos((a-b)x)+12cos((a+b)x)cos(ax)cos(bx)=12cos((a-b)x)+12cos((a+b)x) These formulas may be derived from the sum-of-angle formulas for sine and cosine. Evaluate sin(5x)cos(3x)dx.sin(5x)cos(3x)dx. Apply the identity sin(5x)cos(3x)=12sin(2x)-12sin(8x).sin(5x)cos(3x)=12sin(2x)-12sin(8x). Thus, sin(5x)cos(3x)dx=12sin(2x)dx+12sin(8x)dx=-14cos(2x)-116cos(8x)+C.sin(5x)cos(3x)dx=12sin(2x)dx+12sin(8x)dx=-14cos(2x)-116cos(8x)+C. Evaluate cos(6x)cos(5x)dx.cos(6x)cos(5x)dx. Before discussing the integration of products and powers of tanxtanx and secx,secx, it is useful to recall the integrals involving tanxtanx and secxsecx we have already learned: sec2xdx=tanx+Csec2xdx=tanx+C secxtanxdx=secx+Csecxtanxdx=secx+C tanxdx=ln|secx|+Ctanxdx=ln|secx|+C secxdx=ln|secx+tanx|+C.secxdx=ln|secx+tanx|+C. For most integrals of products and powers of tanxtanx and secx,secx, we rewrite the expression we wish to integrate as the sum or difference of integrals of the form tanjxsec2xdxtanjxsec2xdx or secjxtanxdx.secjxtanxdx. As we see in the following example, we can evaluate these new integrals by using u-substitution. Evaluate sec5xtanxdx.sec5xtanxdx. Start by rewriting sec5xtanxsec5xtanx as sec4xsecxtanx.sec4xsecxtanx. sec5xtanxdx=sec4xsecxtanxdxLetu=secx;then,du=secxtanxdx.=u4duEvaluate the integral.=15u5+CSubstitutesecx=u.=15sec5x+Csec5xtanxdx=sec4xsecxtanxdxLetu=secx;then,du=secxtanxdx.=u4duEvaluate the integral.=15u5+CSubstitutesecx=u.=15sec5x+C You can read some interesting information at this website to learn about a common integral involving the secant. Evaluate tan5xsec2xdx.tan5xsec2xdx. We now take a look at the various strategies for integrating products and powers of secxsecx and tanx.tanx. To integrate tankxsecjxdx,tankxsecjxdx, use the following strategies: If jj is even and j2,j2, rewrite secjx=secj-2xsec2xsecjx=secj-2xsec2x and use sec2x=tan2x+1sec2x=tan2x+1 to rewrite secj-2xsecj-2x in terms of tanx.tanx. Let u=tanxu=tanx and du=sec2xdx.du=sec2xdx. If kk is odd and j1,j1, rewrite tankxsecjx=tank-1xsecj-1xsecxtanxtankxsecjx=tank-1xsecj-1xsecxtanx and use tan2x=sec2x-1tan2x=sec2x-1 to rewrite tank-1xtank-1x in terms of secx.secx. Let u=secxu=secx and du=secxtanxdx.du=secxtanxdx. (Note: If jj is even and kk is odd, then either strategy 1 or strategy 2 may be used.) If kk is odd where k3k3 and j=0,j=0, rewrite tankx=tank-2xtan2x=tank-2x(sec2x-1)=tank-2xsec2x-tank-2x.tankx=tank-2xtan2x=tank-2x(sec2x-1)=tank-2xsec2x-tank-2x. It may be necessary to repeat this process on the tank-2xtank-2x term. If kk is even and jj is odd, then use tan2x=sec2x-1tan2x=sec2x-1 to express tankxtankx in terms of secx.secx. Use integration by parts to integrate odd powers of secx.secx. Evaluate tan6xsec4xdx.tan6xsec4xdx. Since the power on secxsecx is even, rewrite sec4x=sec2xsec2xsec4x=sec2xsec2x and use sec2x=tan2x+1sec2x=tan2x+1 to rewrite the first sec2xsec2x in terms of tanx.tanx. Thus, tan6xsec4xdx=tan6x(tan2x+1)sec2xdxLetu=tanxanddu=sec2xdx.=u6(u2+1)duExpand.=(u8+u6)duEvaluate the integral.=19u9+17u7+CSubstitutetanx=u.=19tan9x+17tan7x+C.tan6xsec4xdx=tan6x(tan2x+1)sec2xdxLetu=tanxanddu=sec2xdx.=u6(u2+1)duExpand.=(u8+u6)duEvaluate the integral.=19u9+17u7+CSubstitutetanx=u.=19tan9x+17tan7x+C. Evaluate tan5xsec3xdx.tan5xsec3xdx. Since the power on tanxtanx is odd, begin by rewriting tan5xsec3x=tan4xsec2xsecxtanx.tan5xsec3x=tan4xsec2xsecxtanx. Thus, tan5xsec3x=tan4xsec2xsecxtanx.Writetan4x= (tan2x)2.tan5xsec3xdx=(tan2x)2sec2xsecxtanxdxUsetan2x=sec2x-1.=(sec2x-1)2sec2xsecxtanxdxLetu=secxanddu=secxtanxdx.=(u2-1)2u2duExpand.=(u6-2u4+u2)duIntegrate.=17u7-25u5+13u3+CSubstitutesecx=u.=17sec7x-25sec5x+13sec3x+C.tan5xsec3x=tan4xsec2xsecxtanx.Writetan4x= (tan2x)2.tan5xsec3xdx=(tan2x)2sec2xsecxtanxdxUsetan2x=sec2x-1.=(sec2x-1)2sec2xsecxtanxdxLetu=secxanddu=secxtanxdx.=(u2-1)2u2duExpand.=(u6-2u4+u2)duIntegrate.=17u7-25u5+13u3+CSubstitutesecx=u.=17sec7x-25sec5x+13sec3x+C. Evaluate tan3xdx.tan3xdx. Begin by rewriting tan3x=tanxtan2x=tanx(sec2x-1)=tanxsec2x-tanx.tan3x=tanxtan2x=tanx(sec2x-1)=tanxsec2x-tanx. Thus, tan3xdx=(tanxsec2x-tanx)dx=tanxsec2xdx-tanxdx=12tan2x-ln|secx|+C.tan3xdx=(tanxsec2x-tanx)dx=tanxsec2xdx-tanxdx=12tan2x-ln|secx|+C. For the first integral, use the substitution u=tanx.u=tanx. For the second integral, use the formula. Integrate sec3xdx.sec3xdx. This integral requires integration by parts. To begin, let u=secxu=secx and dv=sec2xdx.dv=sec2xdx. These choices make du=secxtanxdu=secxtanx and v=tanx.v=tanx. Thus, sec3xdx=secxtanx-tanxsecxtanxdx=secxtanx-tan2xsecxdxSimplify.=secxtanx-(sec2x-1)secxdxSubstitutetan2x=sec2x-1.=secxtanx+secxdx-sec3xdxRewrite.=secxtanx+ln|secx+tanx| -sec3xdx.Evaluatesecxdx.sec3xdx=secxtanx-tanxsecxtanxdx=secxtanx-tan2xsecxdxSimplify.=secxtanx-(sec2x-1)secxdxSubstitutetan2x=sec2x-1.=secxtanx+secxdx-sec3xdxRewrite.=secxtanx+ln|secx+tanx|-sec3xdx.Evaluatesecxdx. We now have sec3xdx=secxtanx+ln|secx+tanx|-sec3xdx.sec3xdx=secxtanx+ln|secx+tanx|-sec3xdx. Since the integral sec3xdxsec3xdx has reappeared on the right-hand side, we can solve for sec3xdxsec3xdx by adding it to both sides. In doing so, we obtain 2sec3xdx=secxtanx+ln|secx+tanx|.2sec3xdx=secxtanx+ln|secx+tanx|. Dividing by 2, we arrive at sec3xdx=12secxtanx+12ln|secx+tanx|+C.sec3xdx=12secxtanx+12ln|secx+tanx|+C. Evaluate tan3xsec7xdx.tan3xsec7xdx. Evaluating secnxdxsecnxdx for values of nn where nn is odd requires integration by parts. In addition, we must also know the value of secn-2xdxsecn-2xdx to evaluate secnxdx.secnxdx. The evaluation of tannxdxtannxdx also requires being able to integrate tann-2xdx.tann-2xdx. To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of secxsecx or tanxtanx with the integral of a lower power of secxsecx or tanx.tanx. secnxdx=1n-1secn-2xtanx+n-2n-1secn-2xdxsecnxdx=1n-1secn-2xtanx+n-2n-1secn-2xdx tannxdx=1n-1tann-1x-tann-2xdxtannxdx=1n-1tann-1x-tann-2xdx The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of tanx.tanx. Apply a reduction formula to evaluate sec3xdx.sec3xdx. By applying the first reduction formula, we obtain sec3xdx=12secxtanx+12secxdx=12secxtanx+12ln|secx+tanx|+C.sec3xdx=12secxtanx+12secxdx=12secxtanx+12ln|secx+tanx|+C. Evaluate tan4xdx.tan4xdx. Applying the reduction formula for tan4xdxtan4xdx we have tan4xdx=13tan3x-tan2xdx=13tan3x-(tanx-tan0xdx)Apply the reduction formula totan2xdx.=13tan3x-tanx+1dxSimplify.=13tan3x-tanx+x+C.Evaluate1dx.tan4xdx=13tan3x-tan2xdx=13tan3x-(tanx-tan0xdx)Apply the reduction formula totan2xdx.=13tan3x-tanx+1dxSimplify.=13tan3x-tanx+x+C.Evaluate1dx. Apply the reduction formula to sec5xdx.sec5xdx. Section 3.2 Exercises Fill in the blank to make a true statement. 69. sin2x+_______=1sin2x+_______=1 70. sec2x-1=_______sec2x-1=_______ Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power. 71. sin2x=_______sin2x=_______ 72. cos2x=_______cos2x=_______ Evaluate each of the following integrals by u-substitution. 73. 74. 75. tan5(2x)sec2(2x)dxtan5(2x)sec2(2x)dx 76. sin7(2x)cos(2x)dxsin7(2x)cos(2x)dx 77. tan(x2)sec2(x2)dxtan(x2)sec2(x2)dx 78. tan2xsec2xdxtan2xsec2xdx Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.) 81. 83. sin5xcos2xdxsin5xcos2xdx 84. sin3xcos3xdxsin3xcos3xdx 85. 86. 87. 89. 90. 94. For the following exercises, find a general formula for the integrals. 95. sin2axcosaxdxsin2axcosaxdx 96. sinaxcosaxdx.sinaxcosaxdx. Use the double-angle formulas to evaluate the following integrals. 100. sin2xcos2xdxsin2xcos2xdx 101. sin2xdx+cos2xdxsin2xdx+cos2xdx 102. sin2xcos2(2x)dxsin2xcos2(2x)dx For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. 103. 02cosxsin2xdx02cosxsin2xdx 104. 0sin3xsin5xdx0sin3xsin5xdx 105. 0cos(99x)sin(101x)dx0cos(99x)sin(101x)dx 106. -cos2(3x)dx-cos2(3x)dx 107. 02sinxsin(2x)sin(3x)dx02sinxsin(2x)sin(3x)dx 108. 04cos(x/2)sin(x/2)dx04cos(x/2)sin(x/2)dx 109. /6/3cos3xsinxdx/6/3cos3xsinxdx (Round this answer to three decimal places.) 110. -/3/3sec2x-1dx-/3/3sec2x-1dx 111. 0/21-cos(2x)dx0/21-cos(2x)dx 112. Find the area of the region bounded by the graphs of the equations y=sinx,y=sin3x,x=0,andx=2.y=sinx,y=sin3x,x=0,andx=2. 113. Find the area of the region bounded by the graphs of the equations y=cos2x,y=sin2x,x=-4,andx=4.y=cos2x,y=sin2x,x=-4,andx=4. 114. A particle moves in a straight line with the velocity function v(t)=sin(t)cos2(t).v(t)=sin(t)cos2(t). Find its position function x=f(t)x=f(t) if f(0)=0.f(0)=0. 115. Find the average value of the function f(x)=sin2xcos3xf(x)=sin2xcos3x over the interval [-,].[-,]. For the following exercises, solve the differential equations. 116. dydx=sin2x.dydx=sin2x. The curve passes through point (0,0).(0,0). 117. dyd=sin4()dyd=sin4() 118. Find the length of the curve y=ln(cscx),4x2.y=ln(cscx),4x2. 119. Find the length of the curve y=ln(sinx),3x2.y=ln(sinx),3x2. 120. Find the volume generated by revolving the curve y=cos(3x)y=cos(3x) about the x-axis, 0x36.0x36. For the following exercises, use this information: The inner product of two functions f and g over [a,b][a,b] is defined by f(x)?g(x)=f,g=abf?gdx.f(x)?g(x)=f,g=abf?gdx. Two distinct functions f and g are said to be orthogonal if f,g=0.f,g=0. 121. Show that {sin(2x),cos(3x)}{sin(2x),cos(3x)} are orthogonal over the interval [-,].[-,]. 122. Evaluate -sin(mx)cos(nx)dx.-sin(mx)cos(nx)dx. 123. Integrate y=tanxsec4x.y=tanxsec4x. For each pair of integrals, determine which one is more difficult to evaluate. Explain your reasoning. 124. sin456xcosxdxsin456xcosxdx or sin2xcos2xdxsin2xcos2xdx 125. tan350xsec2xdxtan350xsec2xdx or tan350xsecxdxtan350xsecxdx

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