Calculus

INTEGRATION –II INTEGRATION USING TRIGONOMETRIC IDENTITIES When the integrand involves trigonometric functions which can not be solved by the basic integration formulas, we need to use the different trigonometric identities (together with substitution if needed) to get them into a form in which the basic integration formula scan be applied. The trigonometric identities often used for integration are

1f 1f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f sec t = f csc t = f cot t = f cos t sin t tan t 2 sin t + cos 2 t = 1 tan2 t + 1 = sec 2 t cot 2 t + 1 = csc 2 t 1f @ cos 2α 2 f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f sin α = 2 1f +f cos 2α f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f cos 2 α = f 2 1f f sinαcosα = f sin 2α 2 D b c b cE 1f f f sin α + β + sin α @ β sinαcosβ = 2 D b c b cE 1f f cosαsinβ = f sin α + β @ sin α @ β 2 D b c b cE 1f f f cosαcosβ = cos α + β + cos α @ β 2 D b c b cE 1f f f sinαsinβ = cos α @ β @ cos α + β 2

sin tf f f f f f f f f f f f f tan t = f cos t

3 Example : Evaluate Z sin x cos 3 x dx

Solution : Z sin3 x cos 3 x dx = Z sin3 x cos 2 x A cos xdx

= Z sin x 1 @sin x A cos x dx let sin x = t, cos x dx = dt 3

b

2

c

B

= Z t 3 @ t 5 dt b

c

1f 1f f f f4 f f t @ t6 + C 6 4 1f 1f 4 f f f f f 6 = sin x @ sin x + C 4 6 =

Example : Evaluate Z sec 6 x dx Solution : Z sec 6 x dx = Z sec 4 x Asec 2 x dx c2

= Z sec 2 x Asec 2 x dx b

c2

= Z tan2 x + 1 Asec 2 x dx b

B

let tan x = t [ sec 2 x dx = dt

c2

= Z t 2 + 1 A dt b

= Z t 4 + 2t 2 + 1 dt b

=

c

1f 2f 1f f f5 2f f f f f 5 f f t + t3 + t = tan x + tan3 x + tan x + C 3 5 3 5

C

C

TRIGONOMETRIC SUBSTITUTION The following two substitutions are useful in many cases : 1 A For Z sin x cos n x dx : If m is odd, we substitute u = cos x A If n is odd we substitute m

u = sin x A

2 A For Z tanm x sec n x dx : If m is odd, we substitute u = sec x A If n is even we substitute u = tan x A

Some more integrations may be simplified by the following substitution : w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

1 A For the integrand in the form q a 2 @ x 2 , substitute x = a sin t w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

2 A For the integrand in the form q a 2 + x 2 , substitute x = a tan t w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

3 A For the integrand in the form q x 2 @ a 2 , substitute x = a sec t w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w 2 2 2 2 2 2 q 2 q 2 q For the integrands in the forms a @ b x , a + b x and b x @ a 2 ,

we have to substitute x =

af af af f f f f f f f f f sin t, x = tan t, x = sec t respectively A b b b

The above substitutions have been used in making the follwoing formulas: w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w wM L M L 1f 1f fq 2 f f 2M q 2 Z q x 2 + a 2 dx = f x x + a 2 + a 2 lnL Lx + x + a M + C 2 2 w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w wM

L 1f 1f fq 2 f f 2M q 2 M Z q x 2 @ a 2 dx = f x x @ a 2 @ a 2 lnL Lx + x @ a M + C

2

2

L

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w 1f 1f xf f f f2 f f f 2 2 2 q q Z a @ x dx = x a @ x 2 + f a arcsin + C

2

2

a

3 Example : Evaluate Z sin x cos 4 x dx

Solution : Z sin3 x cos 4 x dx

=Z sin x cos 4 x sin x dx 2

=Z 1 @cos 2 x cos 4 x sin x dx b

c

=Z 1 @ t 2 t 4 @ dt b

c

`

taking cos x = t, sinx dx = @ dt

a

1f 1f f f f7 f f t @ t5 + C 7 5 1f 1f f f f f f = cos 7 x @ cos 5 x + C 5 7

=

Example : Evaluate Z tan3 x sec 4 x dx Solution : Z tan3 x sec 4 x dx

=Z tan2 x sec 3 x A sec x tan x A dx =Z t 2 + 1 t 3 dt b

c

1f 1f f f f f f = t5 + t4 + C 5 4 1f 1f f f f f f = sec 5 x + sec 4 x + C 4 5

taking sec x = t, sec x tan x dx = dt

dx f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w Example : Evaluate Z f 2q 2 x 4@x

xf f f Solution : Taking a right triangle with one acute angle t such that sin t = f 2 x

2

t w w w w w w w w w w w w w w w w w w w w w w w w w w w w w q4 @ x 2

dx dx f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Z f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w=Z w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w x 2 q4 @ x 2

x 2 q2 @ x 2 2

2f cos tf dt f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f =Z f 2 4 sin t A 2 cos t 1f f f Z cosec 2 t dt = f 4 1f f f =@ f cot t + C 4 =@

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w q4 @ x 2 f f f f f f f f f f f f f f f f f f f f f f f f f f f f

4x

+ C

B

taking x = 2 sin t, dx = 2 cos t dt

C

INTEGRATION USING PARTIAL FRACTIONS ` a

Pf xf f f f f f f f f f f f f f f ` a where A rational function is defined as the ratio of two polynomials in the form Q x

P(x) and Q(x) are polynomials in x and Q x ≠ 0. If the degree of P(x) is less than the ` a

degree of Q(x), then the rational function is called proper, otherwise, it is called improper. Like the case of improper fractions reducible to proper fractions, improper

rational functions can be reduced to proper rational functions by long division process. ` a

` a

` a

xf Pf xf Pf xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f ` a Pf f f f f f f f f f f f f f f f f ` a is improper, then ` a = T x + 1` a where T(x) is a polynomial in x Thus, if Q x Q x Q x ` a

3 Pf xf xf xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 1f ` a is a proper rational fraction. For example , 2 and f . =x + f 2 Q x x @1 x @1

The rational fractions, which we shall consider here for integration, will be those whose denominator can be factored into linear and quadratic factors. We shall write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. ` a

` a

Pf xf Pf xf f f f f f f f f f f f f f f f f f f f f f f f f f f f ` a dx , where ` a is a proper rational function. Assume that we want to evaluate Z f Q x Q x

We can write the integrand as a sum of simpler rational function as per some of the examples in the following table : Form of the rational function px + qf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f ` a` a,a ≠b x @a x @b px + qf f f f f f f f f f f f f f f f f f f f f f f f f `

x @a

a2

px +f qx + rf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f ` a` a` a,a ≠b ≠c 2

x @a x @b x @c

px + qx +f rf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f ` a` a2 , a ≠ b 2

x @a x @b

Form of the partial fractions A B f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f + x @a x @b A B f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f +` a 2 x @a x @a B C A f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

x @a

+

x @b

+

x @c

B C A f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f + +` a2 x @a x @b x @b

px +f qx + rf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f c ` ab 2

A Bx +f C f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f F where x 2 + bx + c

2

x @ a x + bx + c

x @a

+

x 2 + bx + c

xf +f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Example : Evaluate Z f dx 2 x @ 5x + 6 2

Solution : By long division we get 2 +f 1f 5x @ 5f 5x @ 5f xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f a` a =1+ f = 1 + `f 2 2 x @2 x @3 x @ 5x + 6 x @ 5x + 6 5x @ 5f A B f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f a` a = Let `f + f x @2 x @3 x @2 x @3 ` a ` a 5x @ 5 = A x @ 3 + B x @ 2 For x = 2, @ A = 5, so A = @ 5 For x = 3, B = 10 2 xf +f 1f 5f 10 f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Thus f =1@ f + f 2 x @2 x @3 x @ 5x + 6 xf +f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Z f dx 2 2

x @ 5x + 6

1f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f dx + 10 Z f dx = Z dx @ 5 Z f x @2 x @3 = x @ 5 ln |x @ 2| + 10 ln |x @ 3| + C

cannot be factored further

G

dx f f f f f f f f f f f f f f f f f f Example : Evaluate Z f 4 x @1 Solution : 1f 1f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f cb c=` c = bf a` ab 4 x2@ 1 x2 + 1

x @1

x @ 1 x + 1 x2 + 1

1f A B + D f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Cx f f f f f f f f f f f f f f f f f f f f f f c= Let `f + f + f a` ab 2 2 x @1 x + 1 x + 1 x @1 x + 1 x + 1

1 = A x + 1 x 2 + 1 + B x @ 1 x 2 + 1 + Cx + D x @ 1 x + 1 `

For x = 1,

ab

For x = @ 1, For x = 0, For x = 2,

c

ab

`

1f f f 1 = 4A, A = f 4

c `

a`

a

1f f f 1 = @ 4B, B = @ f 4 f g 1f 1f 1f f f f f f 1 = A @B@D = f @ @ f @ D, D = @ f 4 4 2 15 1f f f f f f f 5f f f f 1 = 15A + 5B + 3 2C + D = f @ f + 3 2C @ f = 6C + 1, C = 0 4 4 2 `

a

1f 1f 1f 1f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Z f Z f Z f dx = f dx @ f dx 4 x @1

a`

4 x @1

4 x+1

f

d

e 1f f f f dx 2f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f +Z 2

0A x @

x +1

1f 1f 1f dx f f f f f f f f f f f f f f f f f f f f f f Z f = f ln|x @ 1| @ f ln|x + 1| @ f 2 4 4 2 x +1 1f 1f 1f f f f f f = f ln|x @ 1| @ f ln|x + 1| @ f arctan x + C 4 L 4 2 M M 1f @ 1f f f fL f f f f f f f f f f f f f f 1f f f Lxf M M@ arctan x + C = lnL L M 4 x+1 2

g

INTEGRATION OF HYPERBOLIC FUNCTIONS The following formulas are used for hyperbolic functions : Z sinh x dx = cosh x + C Z cosh x dx = sinh x + C Z tanh x dx = ln |cosh x| + C Z coth x dx = ln |sinh x| + C

Z sech 2 x dx = tanh x + C Z csch 2 x dx = @ coth x + C Z sech x tanh x dx = @ sech x + C Z csch x coth x dx = @ csch x + C

dx @ 1 xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Z f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w = sinh +C a

qx 2 + a 2

dx @ 1 xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Z f w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w = cosh + C, x>a>0 a

qx 2 @ a 2

dx f f f f f f f f f f f f f f f f f f f f f 1f f f f @ 1 xf f f f Z f = tanh + C, 2 2 a @x

a

a

dx 1f f f f f f f f f f f f f f f f f f f f f f f f f @ 1 xf f f f Z f = @ coth + C, 2 2 x @a

a

a

x 2 < a2 x 2 > a2

APPLICATIONS OF INDEFINITE INTEGRALS A. FINDING EQUATIONS OF FAMILY OF CURVES :

If we know the equation y = f x of a curve, the slope of it at any point x, f x given by

b

` a

` a

` ac

will be

f. x . Conversely, when the slope of the curve at any point is given by

dy f f f f f f f ` a m= f = f. x , dx

the

equation

of

the

family

of

curves

will

be

given

by

y =Z f. x dx = f x + C . ` a

` a

To find out a particular curve from the family of curves, we need an initial condition, generally given as the coordinates of one point on the required curve, which we can use to find the arbitrary constant C.

Example : Find the equation of the curve whose slope at any point (x,y) is given by 4x f f f f f f f f and one point on the curve is m =@ 9y

Solution :

f

w w w w w w w w w w w w 3f f f p2 , p2 . What kind of a curve is this? 2

g

4f xf f f f f f m =@ f 9y dy 4f xf f f f f f f f f f f f f f =@ f 9y dx 9y dy = @ 4x dx Z 9y dy = @Z 4x dx 2 2 yf xf f f f f f f f f f f 9 f =@4 f +C 2 2 4x 2 + 9 y 2 = 2C this is the general equation of the family of curve A

w w w w w w w w w w w w 3f f p2 , p2 in this curve we get putting the point f 2

4

f g 9f f f

2

f

g

+ 9 2 = 2C

2C = 36

` a

# The equation of the curve is This is an Ellipse A

4x 2 + 9 y 2 = 36

B. DISTANCE, VELOCITY AND ACCELERATION :

In many applications the distance, velocity and acceleration are functions of time. In applications of derivatives, we have seen that instantaneous velocity is the derivative of the distance function and instantaneous acceleration is the derivative of the velocity

function. As integration is the inverse operation of differentiation, the indefinite integral of the acceleration function represents the velocity function and the indefinite integral of the velocity function represents the distance function. If the distance function, velocity function and the acceletaion function of time are ` a ` a ` a s t , v t and a t respectively ` a ` a ` a ` a and v . t = a t then s. t = v t conversely, Z v t dt = s t + C ` a

` a

and Z a t dt = v t + C ` a

` a

In all the distance-velocity-acceleration problem one side is taken positive, and the opposite is negative. While throwing up a ball, if we take the upward direction positive, then s t ≥ 0 for any point in the motion, v t ≥ 0 in the upward motion and v t ≤ 0 ` a

in the downward motion, downwards.

` a

` a

` a

a t < 0 as the acceleration due to gravity is always

Example : A ball is thrown upward from the ground at a velocity 192 ft/sec2 . When will the velocity be zero? How high will it reach?

Solution : F negative sign as it is always G

We know acceleration due to gravity is @ 32 ft s@ 2 a t = @ 32 ` a

v t =Z a t dt =Z @ 32 dt = @ 32 t + C ` a

` a

putting t = 0 in this, we get ` a ` a v 0 = @ 32 0 + C

192 = C as initial velocity was 192, v 0 = 192 B

` a

so v t = @ 32 t + 192 ` a when the velocity is zero, v t = 0 @ 32 t + 192 = 0 t =6 The velocity will be zero after 6 seconds A ` a

s t =Z v t dt =Z @ 32t + 192 ` a

` a

`

a

= @ 16t 2 + 192t + C1 at t = 0, s 0 = 0, C1 = 0 ` a

s t = @ 16t 2 + 192t As it reaches the highest position at t = 6, ` a

s 6 = @ 16 6 + 192 6 = 576 It will reach a height of 576 ft A ` a

` a2

` a

C

donwards