x n cos 2x dx. dx = nx n 1 and v = 1 2 sin(2x). Andreas Fring (City University London) AS1051 Lecture Autumn / 36

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1 We saw in Example 5.4. that we sometimes need to apply integration by parts several times in the course of a single calculation. Example 5.4.4: For n let S n = x n cos x dx. Find an expression for S n in terms of S n, and hence evaluate S 4. Let u = x n and dv dx Integrating by parts we have = cos x. Then du dx = nx n and v = sin(x). Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36 x n cos x dx = x n n sin(x) x n sin x dx = x n sin(x) + n 4 x n cos x n(n ) x n cos x dx 4 = x n sin(x) + n 4 x n cos x n(n ) S n 4 where the second equality follows using integration by parts with u = n x n and dv dx = sin x. Thus we have found a formula for S n in terms of S n. Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36

2 Clearly S = cos x dx = sin x + C. Hence S = x sin(x) + 4 x cos x 4 sin x + C for some constant C and S 4 = x 4 sin(x) x 3 cos x ( x 3 sin(x) + x cos x 4 sin x + C ) for some constant C. = 4 (x 4 6x + 3) sin x + (x 3 3x) cos x + C Andreas Fring (City University London) AS5 Lecture -4 Autumn 3 / 36 In some examples integration by parts does not lead to a simpler integral. However, even in these cases we can sometimes use this method to solve the original problem. Example 5.4.5: Calculate e x cos x dx. Let u = e x and dv dx = cos x. Integrating by parts we obtain e x cos x dx = e x sin x e x sin x dx. Andreas Fring (City University London) AS5 Lecture -4 Autumn 4 / 36

3 Integrating by parts again we have [ e x cos x dx = e x sin x e x cos x + ] e x cos x dx. The final integral is identical to that we first wished to calculate, however we can now rearrange this formula to obtain e x cos x dx = e x sin x + e x cos x + C from which we deduce that e x cos x dx = (ex sin x + e x cos x + C) Andreas Fring (City University London) AS5 Lecture -4 Autumn 5 / The definite integral If g(x) dx = G(x) + C then we define b which we also denote by a g(x) dx = G(b) G(a) [ ] b G(x). a Andreas Fring (City University London) AS5 Lecture -4 Autumn 6 / 36

4 Example 5.5.: 4 [ ] 4 (x + 3) dx = x + 3 = ( 7 ) = Andreas Fring (City University London) AS5 Lecture -4 Autumn 7 / 36 In the next example we will apply Example Example 5.5.: π sin 4 x cos x dx = [ ] π 5 sin5 x = 5 = 5. Andreas Fring (City University London) AS5 Lecture -4 Autumn 8 / 36

5 When integrating a definite integral by substitution we must be careful to convert the limits into the new variable. Example 5.5.3: Calculate 4 x dx. Let x = sin θ, so dx dθ = cos θ. We have 4 x = 4 4 sin θ = 4 cos θ and in changing variable we have x = θ = x = θ = π. Andreas Fring (City University London) AS5 Lecture -4 Autumn 9 / 36 Combining these observations we obtain π 4 x dx = = = π π [ cos θ cos θ dθ 4 cos θ dθ ( + cos θ) dθ (θ + sin θ )] π = [ π ] = + = π. Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36

6 Lecture 5.6 Integration as a measure of content y=f(x) The area contained between the curve y = f (x), the lines x = a and x = b (for a < b) and the x-axis is given by a b b a f (x) dx. Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36 This follows from the definition of integration as a measure: f(x k ) x k xk δ area n f (x k )δx k k= and the fundamental theorem of calculus which states that this definition agrees with that coming from the antiderivative. Note that this result relies on the convention that area below the x-axis is negative. When calculating area we do not use this convention, so the answer will have to be adjusted appropriately. Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36

7 So for a b c we have b a f (x) dx = c although the total area is clearly non-zero. If b < a we define b a f (x) dx = b a b f (x) dx f (x) dx. (This must clearly be the case from the definition of integration using the antiderivative.) Andreas Fring (City University London) AS5 Lecture -4 Autumn 3 / 36 Example 5.6.: Find the area contained between the quadratic y = 3 + x x and the x-axis. We have y = (3 x)(x + ), and from the graph we see that 3 area = = x x dx [3x + x x 3 3 ] 3 = 3 3. Andreas Fring (City University London) AS5 Lecture -4 Autumn 4 / 36

8 Example 5.6.: Find the area contained in the third arc of the curve for x. y = x sin x π π 3π area = 3π π x sin x dx (use integration by parts) [ ] 3π [ ] 3π = x cos x ( cos x) dx = x cos x + sin x π π = [( 3π)( ) + ] [( π)() + ] = 5π. Andreas Fring (City University London) AS5 Lecture -4 Autumn 5 / 36 Note that if the example had asked for the second and third arcs, we would have calculated 3π π x sin x dx x sin x dx. π Example 5.6.3: Find the area enclosed by the line y = x and the curve y = x 3 3x. π Line and curve intersect when i.e. when x 3 3x = x x(x + )(x ) =..5 A B Andreas Fring (City University London) AS5 Lecture -4 Autumn 6 / 36

9 Let y = x and y = x 3 3x. Then area A = y y dx = = [ x 4 x 3 3x x dx x 3 x ] = 3 3. and area B = y y dx = Therefore the total area is A + B = 3 3. x 3 + 3x + x dx = [ x 4 + x 3 + x ] = 4. Andreas Fring (City University London) AS5 Lecture -4 Autumn 7 / 36 Lecture 3 6. Real functions II 6. Inverse trigonometric functions We would like to define the inverse of sin, cos, and tan, to be denoted sin, cos, and tan. Note: (i) For these to be functions we have to restrict the range. (ii) sin y does not mean (sin y). This is an unfortunate problem with using sin n y = (sin y) n. If n = we must not do this! Andreas Fring (City University London) AS5 Lecture -4 Autumn 8 / 36

10 Function Domain Range Definition y = sin x x π y π x = sin y y = cos x x y π x = cos y y = tan x R π < y < π x = tan y Note that sin and tan are increasing, odd functions, while cos is decreasing. Sometimes we write arcsin x for sin x and similarly arccos x for cos x and arctan x for tan x. Andreas Fring (City University London) AS5 Lecture -4 Autumn 9 / 36 The graphs of these functions are: y = sin θ π / y = cos θ π π/ π / Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36

11 y = tan θ π / π / Example 6..: α = sin ( ) implies that sin α = and π α π. Hence α = π 6. Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36 Example 6..: Express sin( cos x) in terms of x only. Let y = cos x. Then sin( cos x) = sin y = sin y cos y. Now cos x = y gives cos y = x with y π, and sin y = cos y = x. Note that sin y as y π, and so sin y = x. Therefore sin( cos x) = x x. Andreas Fring (City University London) AS5 Lecture -4 Autumn / 36

12 Proposition 6..3: We have for x that cos x = π sin x. Proof: Let y = sin x. Then x = sin y with π y π, and x = cos( π π y π. Therefore y) where cos x = π y = π sin x. Andreas Fring (City University London) AS5 Lecture -4 Autumn 3 / 36 Proposition 6..4: We have tan a + tan b = tan ( a + b ab ) + pπ where p = if π < tan a + tan b < π if π < tan a + tan b < π if π < tan a + tan b < π. Proof: Let α = tan a and β = tan b, so π < α, β < π and tan α = a and tan β = b. We have a + b ab = tan α + tan β tan α tan β = tan(α + β) = tan(α + β + nπ) ( ) (for all n Z) and π < α + β < π. Now tan a+b ab must lie between π and π, and equal α + β + nπ, for some value of n. The result now follows by inspection. Andreas Fring (City University London) AS5 Lecture -4 Autumn 4 / 36

13 Example 6..5: Find u such that tan tan 5 = tan u. Let α = tan 3 4, so tan α = 3 4 with π < α < π 5. Let β = tan, so tan β = 5 with π < β < π. Clearly < α, β < π 4 and so < α + β < π. Hence by the last Proposition we have ( tan tan = tan 4 + ) = tan 33. Andreas Fring (City University London) AS5 Lecture -4 Autumn 5 / 36 Example 6..6: Simplify for x. tan x + tan x + x First suppose that x, i.e tan x π 4. Then x +x = + x +x and so +x ; i.e. Hence for x we have tan x + x π 4. tan x + tan x + x π. Andreas Fring (City University London) AS5 Lecture -4 Autumn 6 / 36

14 Now suppose that x >, i.e π 4 < tan x < π x. Then < +x π x 4 < tan +x <. Hence for x > we have <, so Thus for all x we have tan x + tan x + x = tan < tan x + tan x + x < π. x + x +x x ( x +x ) = tan ( x + + x ) = tan () = π 4. Andreas Fring (City University London) AS5 Lecture -4 Autumn 7 / 36 Lecture 4 6. Differentiation of inverse trigonometric functions Let y = sin x. By definition x = sin y with π y π. We differentiate with respect to x: cos y dy dx = so dy dx = cos y. Now cos y = sin y and π y π, hence cos y = + sin y. Thus we have shown that d dx (sin x) = x. Andreas Fring (City University London) AS5 Lecture -4 Autumn 8 / 36

15 Let y = cos x. Then cos x = π sin x and hence d dx (cos x) = x. Finally let y = tan x. By definition x = tan y with π < y < π. We differentiate with respect to x: sec y dy dx = so dy dx = sec y. Now sec y = + tan y and so sec y = + x. Thus we have shown that d dx (tan x) = + x. Andreas Fring (City University London) AS5 Lecture -4 Autumn 9 / 36 Example 6..: Differentiate sin ( x). Let y = sin u with u = x, so du dx = x. Then dy du = u and dy dx = dy du du dx = x x = x( x). Example 6..: Differentiate tan (x + ). Let y = tan (x + ). Then dy dx = + (x + ) = 4x + 4x + = x + x +. Andreas Fring (City University London) AS5 Lecture -4 Autumn 3 / 36

16 6.3 Integration and inverse trigonometric functions First suppose that y = sin (x/a). Then dy dx =. ( x a ) a = a x. Hence ( a x dx = x ) sin + C. a Andreas Fring (City University London) AS5 Lecture -4 Autumn 3 / 36 Next suppose that y = tan (x/a). Then dy dx = + ( x. a ) a = a a + x. Hence x + a dx = ( x ) a tan + C. a We can now integrate rational functions with quadratic denominators. Andreas Fring (City University London) AS5 Lecture -4 Autumn 3 / 36

17 Example 6.3.: Integrate x + x + 5 dx. The denominator does not factorise, so we complete the square. x + x + 5 dx = (x + ) + 4 dx = ( ) x + tan + C. Andreas Fring (City University London) AS5 Lecture -4 Autumn 33 / 36 Example 6.3.: Integrate x + 3 x + x + 5 dx. Note that d dx (x + x + 5) = x +. Thus x + 3 x + x + 5 dx = (x + ) + x + x + 5 dx = x + x + x + 5 dx + = ln(x + x + 5) + tan ( x + (x + ) + 4 dx ) + C. Andreas Fring (City University London) AS5 Lecture -4 Autumn 34 / 36

18 Example 6.3.3: x + x + dx = (x + x + ) dx = (x + ) + dx 4 ( ) ( ) = x + tan + C (Compare with Ex 6...) = tan (x + ) + C. Andreas Fring (City University London) AS5 Lecture -4 Autumn 35 / 36 We can also deal with more complicated rational functions by using these methods together with partial fractions. Finally, we consider the integrals of inverse trigonometric functions. To integrate sin x we use integration by parts with u = sin x and v = x. sin x = x sin x x dx = x x sin x + x + C. Similarly tan x = x tan x x x + dx = x tan x ln(x + ) + C. Andreas Fring (City University London) AS5 Lecture -4 Autumn 36 / 36