Question

Evaluate the integral.

2	1 t3 t2 − 1 dt
√2

Part 1 of 7

Since

2	1 t3 t2 − 1 dt
√2

contains the expression

t² − 1,

we will make the substitution

t = sec θ.

With this, we getdt =

$$sec(θ)tan(θ)

dθ.

Part 2 of 7

Using

t = sec θ,

we can also say that

t2 − 1	=	sec2θ − 21 1
	=	tan θ.

Part 3 of 7

Using

t = sec θ,

we must also convert the integration limits.

If

t = 2,

then

θ	=	sec−1(2)
	=	$$π3​

and if t =

2

, then

θ	=	sec−1 2
	=	$$π4​ .

Part 4 of 7

Using

t = sec θ,

we get

2	1 t3 t2 − 1 dt
√2

=

π/3	51 1 sec3θ tan θ sec θ tan θ dθ
π/4

.This can be further simplified to

π/3	61 1 sec2θ dθ
π/4

,or simply

π/3	cos2θ dθ.
π/4

Part 5 of 7

Since we have an even power of cos x, to evaluate

π/3	cos2θ dθ
π/4

, we must now make the further substitution.cos²θ =

1

2

71

1

+ cos

82

2

θ

Part 6 of 7

We have

π/3 1 2 (1 + cos 2θ) dθ π/4

=

1 2 θ + 1 2 sin 92 2 θ π/3 π/4 .

Part 7 of 7

Now,

1 2 θ + 1 2 sin 2θ π/3 π/4

=

1 2 π 3 + 1 2 sin 2π 10 − 1 2 π 4 + 1 2 sin 2π 11 .

After simplifying we have

2	1 t3 t2 − 1 dt
√2

=

Answer 1