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