Math 231E. Fall 2013. Lecture 8A.Trigonometric Integrals and Trigonometric SubstitutionIn this lecture we will learn how to integrate expressions involving trigonometric functions, orintegrals that can be converted into same.Example 36. Let us consider the integralZsin5(x) cos2(x) dx.Recalling the identitysin2(x) + cos2(x) = 1,we can write sin2(x) = 1 − cos2(x), andsin4(x) = (sin2(x))2= (1 − cos2(x))2= 1 − 2 cos2(x) + cos4(x).Thus we can rewrite our integralZsin5(x) cos2(x) dx =Zsin(x) sin4(x) cos2(x) dx=Zsin(x)(1 − 2 cos2(x) + cos4(x)) cos2(x) dx=Zsin(x)(cos2(x) − 2 cos4(x) + cos6(x)) dx.Now, since we have a sin(x) in the integral, we can writeu = cos(x), du = −sin(x) dx,giving−Z(u2− 2u4+ u6) du = −u33+25u5−u77+ C,orZsin5(x) cos2(x) dx = −cos3(x)3+2 cos5(x)5−cos7(x)7+ C.We can see in the scenario above that the trick has nothing to do with the power of 5 on thesin x, but it uses the fact that it is odd. So, for example, if we consider any integral of the formZsin2k+1(x) cosm(x) dx,where k, m are integers, then we writeZsin2k+1(x) cosm(x) dx =Zsin(x) sin2k(x) cosm(x) dx=Zsin(x)(1 − cos2(x))kcosm(x) dx,and then we make the substitution u = cos(x), du = −sin(x) dx, and we obtainZsin2k+1(x) cosm(x) dx =Z−(1 − u2)kumdu.This is a polynomial that we can always integrate. Similarly, for any integral of the formRcos2k+1(x) sinm(x) dx,we can peel off all but one of the cosines:Zcos2k+1(x) sinm(x) dx =Zcos(x) cos2(x) sinm(x) dx=Zcos(x)(1 − sin2k(x))ksinm(x) dx,and then we make the substitution u = sin(x), du = cos(x) dx, and we obtainZcos2k+1(x) sinm(x) dx =Z(1 − u2)kumdu.Of course, this algorithm will only work when one (or both) of the powers on sines and cosinesare odd. What about an integral of the formZcos2(x) sin4(x) dx?Here we use the “half-angle formulas”sin2(x) =12(1 − cos(2x)), cos2(x) =12(1 + cos(2x)).For example, we can computeZ2π0sin2(x) dx =Z2π012(1 − cos(2x)) dx=Z2π012dx −Z2π012cos(2x) dx= π −12cos(2x)2π0= π −12+12= π.Now, how do we deal with the original problem? We can writeZcos2(x) sin4(x) dx =Z12(1 + cos(2x))12(1 − cos(2x))2dx=18Z(1 + cos(2x))(1 − cos(2x))2dx=18Z(1 − cos(2x) − cos2(2x) + cos3(2x)) dxNow, we can deal with these four integrals separately. The constant is easy, as is the simple cosine.The cube we deal with as we described above, i.e.Zcos3(2x) =Zcos2(2x) cos(2x) dx=Z(1 − sin2(2x)) cos(2x) dx,and make the substitution u = sin(2x), du = 2 cos(2x) dx, to obtainZcos3(2x) =12Z(1 − u2) du =12sin(2x) −16sin3(2x) + C.We use the half-angle again to obtainZcos2(2x) dx =Z12(1 + cos(4x)) =x2+18sin(4x) + C.Putting all of this together givesZcos2(x) sin4(x) dx =x8−116sin(2x) −x16−164sin(4x) +12sin(2x) −16sin3(2x) + C=x16+716sin(2x) −164sin(4x) −16sin3(2x) + C.One of the main reasons these types of trig integrals are useful is that they allow us to dealwith square roots of quadratic forms.Example 37.Zx3p1 − x2dx.If we make the substitution x = sin θ, dx = cos θ dθ, then we havep1 − x2=q1 − sin2(θ) = |cos(θ)|.As long as we take θ ∈ (−π/2, π/2), then cos(θ) > 0 and so |cos(θ) = cos(θ). We then haveZx3p1 − x2dx =Zsin3(θ) · cos(θ) · cos(θ) dθ=Zsin3(θ) cos2(θ) dθ.Since we have an odd number of sin terms above, we “peel off” all but one of them and proceed asabove.Similar to the half-angle formulas, there are the “angle-addition” formulas, namely:sin A cos B =12(sin(A − B) + sin(A + B)),sin A sin B =12(cos(A − B) − cos(A + B)),cos A cos B =12(cos(A − B) + cos(A + B)).In fact, you can see that these addition formulas actually give the half-angle formulas: ChooseA = B in either of the last two lines, and we recover the half-angle formulas.This formula can be useful in integrals of the following form:Example 38. ConsiderZsin(3x) cos(4x) dx.Using the first of the two formulas givessin(3x) cos(4x) =12(sin(3x − 4x) + sin(3x + 4x)) =12(sin(−x) + sin(7x)) =12(sin(7x) − sin(x)).Then we haveZsin(3x) cos(4x) dx =12Z(sin(7x) − sin(x)) dx = −114cos(7x) +12cos(x) + C.We should not think of the half-angle formulas, or the angle addition formulas, as “somethingto memorize”. This is because we know complex numbers and Euler’s Formula! Recall Euler’sformula:eiθ= cos(θ) + i sin(θ).This means that we haveei(A+B)= cos(A + B) + i sin(A + B).We also have, from the rules of multiplying exponents,ei(A+B)= eiAeiB= (cos(A) + i sin(A))(cos(B) + i sin(B)).Multiplying the last two binomials out givescos(A + B) + i sin(A + B) = cos A cos B − sin A sin B + i(cos A sin B + sin A cos B).If two complex expressions are equal, then their real parts, and their imaginary parts, must beequal, so we havecos(A + B) = cos A cos B − sin A sin B,sin(A + B) = cos A sin B + sin A cos B.Now, we can also use parity. Notice thatcos(A−B) = cos A cos(−B)−sin A sin(−B) = cos A cos B−sin A(−sin B) = cos A cos B+sin A sin B.We can then add and substractcos(A − B) + cos(A + B) = 2 cos A cos B,cos(A − B) − cos(A + B) = 2 sin A sin B,recovering two out of three of the laws above. Doing similar manipulations on the sin equationgives the third.We can, finally, use all of these integrals to compute the area of an ellipse.We know that the formula for an ellipse whose two axes are a and b is given by the formulax2a2+y2b2= 1.This means that the upper edge of the ellipse is given by the curvey =rb2−b2a2x2=bapa2− x2.Clearly, we integrate from x = −a to x = a, so the top half of the area isA2=Za−abapa2− x2dx.We should use the substitution x = a sin θ and dx = a cos θ dθ, to obtainZa−abapa2− x2dx =Zπ/2−π/2bapa2− a2sin2θ · a cos θ dθ=Zπ/2−π/2abp1 − sin2θ cos θ dθ=Zπ/2−π/2ab cos2θ dθ= abZπ/2−π/212(1 + cos(2θ)) dθ =abπ2.ThusA2=abπ2,so A =
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