5.7Inverse TrigonometricFunctions: Integration
and Completingthe Square
Ca
uarcaauu
du
Ca
u
aua
du
Ca
u
ua
du
+=−
+=+
+=−
∫
∫
∫
sec1
arctan1
arcsin
22
22
22
Ex.=
−∫ 24 x
dxC
x+
2arcsin
Ex. =+∫ 292 x
dx
€
dx
2( )2
+ 3x( )2
∫ Let u = 3xdu = 3 dx
dxdu
=3
€
=1
3
du
2( )2
+ u( )2
∫ Cx+=
23
arctan23
1
Ex. Integrate by substitution.
∫ −12xe
dx
€
=dx
ex( )2
−1∫ Let u = ex
du = ex dx
dxe
dux=
dxu
du=( ) u
du
u∫
−=
1
12
( )∫−
=12uu
duC
earcC
uarc
x
+=+=1
sec1
sec
Ex. Rewriting the integrand as the sum of two quotients.
dxx
x∫ −
+24
2∫∫ −
+−
= dxx
dxx
x22 4
2
4
Let u = 4 – x2
du = -2x dx
dxx
du=
−2
( )xdu
u
x
2−∫ duu∫−
=−
2
21
2
2 21
−=
u
24 x−−
∫ −=
2222
x
dx
2arcsin2
x=
Final Answer
Cx
x ++−−2
arcsin24 2
Ex. Integrating an improper rational function.
dxx
x∫ +
−423
2
3Do long division and then rewrite theintegrand as the sum of two quotients.
dxx
xx∫ ⎟
⎠
⎞⎜⎝
⎛++
−=4212
3 2
∫∫∫ +−
+−=
4
2
4
123
22 x
dxdx
x
xxdx
∫∫∫ +−
+−=
42
4
263
22 x
dxdx
x
xxdx
( ) Cx
xx
+−+−=2
arctan4ln62
3 22
1-29 odd
Ex. Completing the Square
∫ +− 742 xx
dx
€
=dx
x 2 − 4x + 4( ) + 7 − 4∫
€
=dx
x − 2( )2
+ 3∫ Let u = x – 2
du = dx
€
=du
u( )2
+ 3( )2∫ C
x+
−=
32
arctan31
Ex. Completing the square when the leading coefficient is not 1.
∫ +− 1082 2 xx
dx First, factor out a 1/2
∫ +−=
542
12 xx
dx∫ +−
=542
12 xx
dx
∫ −++−=
45442
12 xx
dx∫ +−
=1)2(2
12x
dx
Let u = x – 2 du = dx ∫ +
=1)(2
12u
duC
x+
−=
1)2(
arctan21
Now complete the squarein the denominator.
Find the area of the region bounded by the graph of
f(x) = , the x-axis, and and 23
1
xx−,
2
3=x .
4
9=x
1 2
2
1
dxxx
∫ −
49
2323
1
dxxx
∫ −
49
2323
1Factor out a neg. inside the rad.
( )dxxx∫
−−=
49
232 3
1
dx
xx∫
+⎟⎠⎞
⎜⎝⎛ +−−
=49
23 2
49
493
1dx
x
∫⎟⎠⎞
⎜⎝⎛ −−⎟
⎠⎞
⎜⎝⎛
=49
2322
23
23
1
( ) 49
2323
23arcsin ⎥
⎦
⎤⎢⎣
⎡ −=
x0arcsin
2
1arcsin −=
60
6
ππ=−=
31-43 odd, 53, 55, 63, 65
Adding and Subtracting Common Denominators
€
2x − 5
x 2 + 2x + 2∫ dx
€
+7
x 2 + 2x + 2∫ dx −
7
x 2 + 2x + 2∫ dx
The derivative of x2 + 2x + 2 is 2x + 2, so to get it, add and subtract 7 over x2 + 2x + 2.Now, put the first two term together.
€
2x + 2
x 2 + 2x + 2∫ dx −
7
x 2 + 2x + 2∫ dx
Now, integrate bothterms. u’/u & arctan
€
2x + 2
x 2 + 2x + 2∫ dx −
7
x 2 + 2x +1( ) + 2 −1∫ dx
€
2x + 2
x 2 + 2x + 2∫ dx −
7
x +1( )2
+12∫ dx
€
=ln x 2 + 2x + 2 − 7 ⋅1
1arctan
x +1
1+C