Download Math 242: Problem 25 of Sec. 8.3 8.3.25. Evaluate the integral

Math 242: Problem 25 of Sec. 8.3
8.3.25. Evaluate the integral
Z
8 dx
(4x2 + 1)2
Solution:
We started this problem correctly in class, but I made a mistake during the computation, so here
is the full solution (with many more details than you need to write, once you are comfortable with
these problems).
The integral is
Z
8 dx
=
(4x2 + 1)2
Z
Z
8 dx
4
x2
+
1
4
2 =
8 dx
1
=
1 2
2
2
16 x + 4
Z
dx
x2
+
.
1 2
4
The substitution rule on the top of page 453 tells us how to deal with this kind of integral. We
identify a2 = 1/4, or a = 1/2, so
1
tan θ.
Then dx =
2
!2 2
1
1
1 2
1
tan θ +
=
tan2 θ +
=
2
4
4
4
Let x =
1 2
x2 +
=
4
1
sec2 θ dθ, and
2
1
1
1
(tan2 θ+1)2 = (sec2 θ)2 =
sec4 θ.
16
16
16
So the integral becomes
Z
Z
Z
Z
Z
dx
(1/2) sec2 θ dθ
16
sec2 θ dθ
dθ
1
1
=
=4
= 4 cos2 θ dθ.
2 =
2
2
(1/16) sec4 θ
2·2
sec4 θ
sec2 θ
x2 + 1
4
To evaluate
R
cos2 θ dθ, use the trigonometric identity
Z
Z
1 + cos 2θ
1 + cos 2θ
2
2
cos θ =
to write 4 cos θ dθ = 4
dθ,
2
2
which is equal to
Z
2
1 + cos 2θ dθ = 2θ + sin 2θ + C = 2θ + 2 sin θ cos θ + C.
To get the last equality, recall the two formulas I suggested you memorize:
sin(A + B) = sin A cos B + sin B cos A
and
cos(A + B) = cos A cos B − sin A sin B.
If we let A = B = θ in the first of these, we get the “double angle” formula
sin 2θ = 2 sin θ cos θ.
continued on next page...
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Math 242: Problem 25 of Sec. 8.3
To finish the solution, we must write 2θ + 2 sin θ cos θ + C in terms of x, using our substitution,
x = 12 tan θ. First, to replace θ, we use
x=
1
tan θ
2
⇔
2x = tan θ
⇔
tan−1 (2x) = θ.
So 2θ = 2 tan−1 (2x).
Finally, to write 2 sin θ cos θ in terms of x, draw a triangle. Since tangent is opposite over
adjacent, and tan θ = 2x = 2x
1 , label the opposite edge 2x and label the adjacent edge 1. Then the
√
hypotenuse must be 4x2 + 1, by the Pythagorean theorem. Since sine is opposite over hypotenuse,
sin θ = √4x2x2 +1 . Cosine is adjacent over hypotenuse, so cos θ = √4x12 +1 . Putting it all together,
2x
1
4x
√
2θ + 2 sin θ cos θ + C = 2 tan−1 (2x) + 2 √
+ C.
+ C = 2 tan−1 (2x) + 2
2
2
4x + 1
4x + 1 4x + 1
Therefore, the answer is
Z
8 dx
4x
= 2 tan−1 (2x) + 2
+ C.
(4x2 + 1)2
4x + 1
Incidentally, you can check this answer using Maxima. Use the Calculus menu, select Integrate and
enter the expression 8/(4*x^2 + 1)^2 (don’t forget the *!). Maxima says the answer is
atan (2 x)
x
8
+
,
4
8 x2 + 2
which, except for the constant C, is the same as our answer.
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