Page 13 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes Thus, we have
v 3 v 5
3
2
 x 4 x dx   32  32  . C

3 5
This will not answer completely the problem since the answer should be given as a function of x.
2

As v  cos   t  1 sin 2   t  1   /2x  , we obtain after easy simplifications.
3
2
Example: Evaluate 0  x  6xdx
Solution:
2
x
First let us complete the square for x  6 . We obtain
2
x  6x  x   3 2  9
which recommends the secant-substitution x + 3 = sec(t). So we have dx = 3 sec (t) tan (t) and
2
x + 6x = 9 (sec (t)– 1) = 9 tan (t). Observe that for x =0, we have sec(t) = 1 which provides t = 0 and
2
2
for x = 3, we have sec(t) = 2 which provides t = /3 . So, we have
3 2  /2  /3 2
0  x  6xdx  0  3tan  3sect  tant   t dt  9 0  tan  sect   .t dt
By means of the trigonometric identities, we obtain

 /3 2  /3 2
0  tan  sect   t dt  0  sec   sect    .t dt
The technique of integration connected to the powers of the secant-function provides
 sec   t dt  In sec   tant    t  C and

1 1
 sec 3   t dt  sec  tant   t  In sec   tant    t  C
2 2
which entails

 /3 3  1 1   /3
0  sec   sect   dt    2 sec  tant   t  2 In sec   tant    t  0  .
t
One would ensure easily that

3 2  /3 2 1
0  x  6xdx  9 0  tan  sect   t dt  3  2 In  2 + 3 . 

Useful trigonometric identities:

1 sin 2   t  cos 2   t


1 tan 2   t  sec 2   t
sec 2   1t   tan 2   t

1
Example: Evaluate  dx .

1 x 2
Solution:
Observe that we cannot use a u-substitution.

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