So,
∫
4
√t
1
2
–
=
dt
√t
2/3
EXAMPLE 17 Evaluate the integral
-2
∫ -5
x4 – 1 x2 + 1
dx
Solution We simplify the integral:
-2
∫ -5
x4 – 1 x2 + 1
dx
-2 (x2 – 1)(x2 + 1)
∫ -5
=
x2 + 1
dx
Which becomes
∫
-2
(x
−5
2
– 1) dx
The integrand is f(x) = x2 – 1, and its antiderivative is F(x) = (x3/3) – x.. This gives
∫
-2 −5
=
(x
– 1) dx
2
[((-2) /3) 3
=
[(–8/3)
=
–2/3
]
–
+ 2
=
(-2)
]
–
–
]
[((-5) /3)
–
3
[(–125/3) =
– (–110/3)
x
(x3/3) –
+ 5
-2 -5
]
(-5)
]
36
Thus,
-2
∫ -5
x4 – 1 x2 + 1
dx
=
36
EXAMPLE 18 Evaluate the integral
∫
π/3 π/6
cosec 2 x
dx