Evaluate the integral. 2 1 t3 t2 − 1 dt √2 Part 1 of 7…
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Question “Evaluate the integral. 2 1 t3 t2 − 1 dt √2 Part 1 of 7…”
Evaluate the integral.
2 |
dt | ||||
√2 |
Part 1 of 7
Since
2 |
dt | ||||
√2 |
contains the expression
t2 − 1,
we will make the substitution
t = sec θ.
With this, we getdt =
$$sec(θ)tan(θ)
Part 2 of 7
Using
t = sec θ,
we can also say that
| = |
| ||||
= | tan θ. |
Part 3 of 7
Using
t = sec θ,
we must also convert the integration limits.
If
t = 2,
then
θ | = | sec−1(2) |
= | $$π3 |
and if t =
2 |
, then
θ | = |
sec−1
| ||||
= | $$π4 |
Part 4 of 7
Using
t = sec θ,
we get
2 |
dt | ||||
√2 |
=
π/3 |
sec θ tan θ dθ | ||
π/4 |
.This can be further simplified to
π/3 |
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 |
1
+ cos
2
θ
Part 6 of 7
We have
| = |
θ +
sin 2 θ
. |
Part 7 of 7
Now,
θ +
sin 2θ
| = |
+
sin
−
+
sin
. |
After simplifying we have
2 |
dt | ||||
√2 |
=
Answer
Conclusion
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