10)

I

sec(r)tan(u)du:

sec(u)

+c

I

(11)

J

csc(u)cot(u)du:

-csc(u)

+

c

EX.

of l0)

J

*r..(*'

)tan(x'

)d*

:

@[

sec(x'

)tan(x'

)

2xdx

;fsec(x'

)

+

c

L2)

J

.r.'udu

:

-cot(u)+c

(13)

J

a'

du

:

a'lln(a)

+c

14)

J

or(r*u'):

tan-'(u)

*c

or

-cot-'u*c

lrJ

du

:

sin-'u+c

^ll

-

"'

EX.

of

14)

J

f.olrl

(l+xn

):

V)[

(zxdxl

[l+(x,

), ]

:

(%)

tmr-'(x,

)

+

c

16)

J

q,

-

sec-'lrrl+.

or-csc-'lrl+c

(inthisproblern,

the%powerisasquareroot)

u1u'-l;"'

Exanrple

of solving

a differential

equations

by the

"variables

separable"

method

or -cos

-'u

*

c

Solve:

dy

:

cos'y

=+

dy

:

dx

sin'x

cos'y

dx

=

sec'Y

dy:

csc'x

dx-

+

.2

smx

NF

J

sec'ydy:

J

csc'xdx

:+

tany:-cotx*c

EX:

A

particle

moves

in

a straight

line

with

acceleration

a:12t2

+6t.It

starts

from

rest

with

an initial

velocity

of

-3

fl/sec.

Find

the

equation

for

distance

s

at time

t.

Since

u--dvld|

we have

the

differential

equation

dv/dt

:

l2t:

+61.

Separating

the

variables

we

ge!

dv:

(12t'z+6t)dt.

tntegrating

both

sides,

J

ou

:

I

gzr'+6t)dt

Which

glves

us, v

:

l\t'13

+

6t'

12

+

s.

Since

v:

-3

when

t:0,

c:

-3.

Substitute

-3 for

c and

we have,

,: 41'+3t'

-3.

since

v:

ds/dt,

we have

another

diff. Eq.,

ds/de4t'+3t'-3.

Separating

the

variables

& integrating,

we

get,

s{n+t'-3t+t.

Since

the

particle

starts from

res!

s:0 when

F0,

so

c:0. Therefore,

the equation

for

distance

is,

s:

ta+t3-31.

NOTE:

When

separating

the variables,

make

sure

the

differentials

are "up

stairs"

and

to

the right

of the

expressions

to

be integrated.