Techniques of Integration

Chapter 8 Techniques of Integration 8. Trigonometric Integrals Summary (a) Integrals of the form sin m x cos n x. () sin k+ x cos n x = ( cos x) k cos n x (sin x ), then apply the substitution u = cos x. () sin m x cos k+ x = sin m x( sin x) k (cos x ), then apply the substitution u = sin x. (3) When both m and n are even, use the following double angle formulas to reduce the degree. (b) Integrals of the form tan m x sec n x. sin x = sin x cos x and cos x = cos x = sin x () tan k+ x sec n x = (sec x ) k sec n x (tan x sec x ), then apply the substitution u = sec x. () tan m x sec k x = tan m x( + tan x) k x (sec x ), then apply the substitution u = tan x. (3) In any cases, tan m x sec n x = sin m x (tan x sec m+n x) and integrate by parts. The formula tan x sec k x = k seck x + C will be helpful in this case. (c) Integrals of the form sin mx cos nx, sin mx sin nx and cos mx cos nx () You can use the formulas sin mx cos nx = (sin(m n)x + sin(m + n)x) sin mx sin nx = (cos(m n)x cos(m + n)x) cos mx cos nx = (cos(m n)x + cos(m + n)x) () Alternatively, you can apply integration by parts twice. (d) Memorizing some special case may be helpful, especially when they are hard to reproduce: sec x = ln sec x + tan x + C and csc x = ln csc x cot x + C.................................................................................................................. Exercise 8..6. Calculate sin x cos x.

8.. Trigonometric Integrals Solution. This corresponds to the case (a)-(). So we use the substitution u = sin x. Then sin x cos x = sin x( sin x) (cos ) = u ( u ) du = 3 u3 u + 7 u7 + C = 3 sin3 x sin x + 7 sin7 x + C. Exercise 8..9. Calculate cos 4 y d y. Solution. This corresponds to the case (a)-(3). cos x = ( + cos x). Applying this, cos 4 y d y = = 4 ( + cos y) d y = 4 + cos y + 8 ( + cos 4 y) d y = = 3 8 x + 4 sin y + 3 sin 4 y + C. By the double angle formula cos x = cos x, we have 4 + cos y + 4 cos y d y 3 8 + cos y + 8 cos 4y d y Exercise 8..34. Calculate tan 3 x sec 3 x. Solution. This corresponds to the case (b)-(). So we use the substitution u = sec x. Then tan 3 x sec 3 x = (sec x ) sec x (tan x sec x ) = (u )u du = u 3 u3 + C = sec x 3 sec3 x + C. Exercise 8..6. Calculate tan x sec x. Solution. This does not correspond to any of (b)-() or (b)-(). So we apply (b)-(3). Letting u = sin x and v = sec x, then tan x sec x = sin x (tan x sec x) = = sin x sec x uv = uv u v sec x = tan x sec x ln sec x + tan x + C.

8.3. Trigonometric Substitution Exercise 8..4. Calculate cos 4x cos 6x. Solution. This corresponds to (c). By using the formula cos 4x cos 6x = (cos x + cos 0x), it follows that cos 4x cos 6x = (cos x + cos 0x) = 4 sin x + 0 sin 0x + C. Alternatively, let us apply integration by parts twice. cos 4x cos 6x = 4 sin 4x cos 6x + 3 sin 4x sin 6x = 4 sin 4x cos 6x 3 8 cos 4x sin 6x + 9 4 cos 4x cos 6x. Solving in terms of the original integral, we get cos 4x cos 6x = 3 sin 4x cos 6x + 0 cos 4x sin 6x + C. You can check that they, though seem different, are actually the same. 8.3 Trigonometric Substitution Summary (a) If the integral involves x + a, then use x = a tan θ. ( = a sec θ dθ and x + a = a sec θ) (b) If the integral involves a x, then use x = a sin θ. ( = a cos θ dθ and a x = a cos θ) (c) If the integral involves x a, then use x = a sec θ. ( = a sec θ tan θ dθ and x a = a tan θ) (d) Complete the square inside the square root if needed.................................................................................................................. Exercise 8.3.. Calculate x 9 x. Solution. This corresponds to the case (b). So we plug x = 3 sin θ. Then x = 9 sin θ dθ = 9 x 9 ( cos θ) dθ 9 x = dθ and hence = 9 θ 9 4 sin θ + C = 9 θ 9 sin θ cos θ + C = 9 sin x x 9 x + C. 3

8.3. Trigonometric Substitution Exercise 8.3.3. Calculate x +. Solution. This corresponds to the case (a). So we use x = tan θ. Then x + = = sec θ dθ = ln sec θ + tan θ + C x + = ln x + x + + C = x ln + x + + C. Here, C = C ln is again an arbitrary constant. Exercise 8.3.. Calculate dz z 3 z 4. Solution. This corresponds to the case (c). So we apply the substitution z = sec θ. Then dz z 3 z 4 = 8 cos θ dθ = sin θ + 3 6 θ + C = sin θ cos θ + 6 6 θ + C = 6 4 z z + 6 arccos + C z = z 4 + 8z 6 arccos + C. z Exercise 8.3.4. Calculate x 4x + 3. Solution. By completing the square, x 4x + 3 = (x ). So by appealing to the case (c), we make the substitution x = sec θ. Then x 4x + 3 = tan θ sec θ dθ. From Exercise 8..6, we know how to integrate this. Thus plugging back, x 4x + 3 = sec θ tan θ ln sec θ + tan θ + C = (x ) x 4x + 3 x ln + x 4x + 3 + C. 4

8.. The Method of Partial Fraction 8. The Method of Partial Fraction Summary How to integrate a rational function P(x)/Q(x)? If P(x)/Q(x) is not proper, apply long division to write P(x)/Q(x) = g(x) + R(x)/Q(x) and integrate separately. If P(x)/Q(x) is proper, factor Q(x) as Q(x) = (x a ) M (x a k ) M k (x + b x + c ) N (x + b l x + c l ) N l. (Here, each factor is different.) Then for each contribution of (x a) M, set up the formula A x a + + A M (x a) M and for each contribution of (x + bx + c) N, set up the formula B x + C x + bx + c + + B N x + C N (x + bx + c). N That is, you will end up equating the following horrible expression: P(x) A, Q(x) = A,M Ak, A k,mk + + + + + + x a (x a ) M x a k (x a k ) M k B, x + C, + + + B,N x + C,N Bl, x + C l, + + + + B l,n l x + C l,nl. x + b x + c (x + b x + c ) N x + b l x + c l (x + b l x + c l ) N l This is called a partial fraction decomposition ) of P(x)/Q(x). Determine the coefficients (either by clearing denominators or by plugging some values of x). Carry out integration!................................................................................................................. Exercise 8..9. Calculate (x )(x 4). Solution. As a special case of partial fraction decomposition, we have (x a)(x b) = b a In this case, we have (x )(x 4) = x 4 = x x a x b. = (ln x 4 ln x ) + C = ln x 4 x + C. Exercise 8..6. Calculate (x + )(x + x). Solution. Note that (x + )(x + x) = (x + ) x. ) Existence and uniqueness of this decomposition is out of the scope of this course. But several analytic and algebraic proof are known.

8.. The Method of Partial Fraction So we can set up the partial fraction decomposition as (x + )(x + x) = A x + B Clearing denominators, x + + C (x + ). = A(x + ) + Bx(x + ) + C x = (A + B)x + (A + B + C)x + A, and comparing both sides, we get A =, B = and C =. This gives (x + )(x + x) = x x + = ln x ln x + + (x + ) x + + C x = ln + x + x + + C. Exercise 8..9. Calculate x (x + ). Solution. By applying the method of partial fraction, we have x (x + ) = x x + = x tan x + C. Exercise 8..43. Calculate x(x + x + ). Solution. We can set up the partial fraction decomposition as follows: x(x + x + ) = A x + B x + C x + x + + B x + C (x + x + ). Clearing denominators, we have = A(x + x + ) + (B x + C )x(x + x + ) + (B x + C )x. Comparing both sides, we get A =, B =, C =, B = and C = 0. Then x(x + x + ) = x x + x + x + (x + ) (x + x + ) = x + x + x + x + (x + ) + (x + x + ) (x + ) + } {{ } (). ((x + ) + ) }{{ } () For (), we apply the substitution u = x + x +. Then du = (x + ) and x + x + x + (x + ) = (x + x + ) u du = u ln u + u + C = ln x + x + + (x + x + ) + C. 6

8.. The Method of Partial Fraction For (), we apply the substitution x + = tan θ. Then (x + ) + = ((x + ) + ) x +x+ = dθ and 8 cos θ dθ = 3 6 θ cos θ sin θ + C 6 = 3 6 θ tan θ 6 sec θ + C = 3 6 arctan Summing up together, we have x(x + x + ) = ln x ln x + x + 3 6 arctan x + x + 8 8 x + x + x + + C x 3 x + x +. Exercise (A challenging integral). Calculate x 4 +. Solution. We give two methods. Method. Notice that x 4 + factors into the following way: x 4 + = (x 4 + x + ) x = (x + ) ( x) = (x + x + )(x x + ). Applying the method of partial fraction, we have x 4 + = x + (x + x + ) x (x x + ) x + = 4 (x + x + ) x 4 (x x + ) + (( x + ) + ) + (( x ) + ). Integrating each term, we get x 4 + = 4 ln(x + x + ) 4 ln(x x + ) + tan ( x + ) + tan ( x ). Method. We rewrite the integral in the following way: x 4 + = x x + x = + x x x + x x + x = + x (x x ) + x (x + x ) Now we apply u = x x and v = x + x separately. Since du = (+ x ) and dv = ( x ), x 4 + = du u + dv = v tan u + ln v + v + C 7

8.. The Method of Partial Fraction Plugging the substitution back, we have x 4 + = x tan + x x 4 ln + x + x + C x + You can check that they are the same (up to constant). 8