8.5. TRIG. SUBSTITUTIONS 201

and so ∫sec3 (u) du =

12[tan(u)sec(u)+ ln |sec(u)+ tan(u)|]+C. (8.6)

Now it follows from 8.5 that in terms of u the set of antiderivatives is given by

34[tan(u)sec(u)+ ln |sec(u)+ tan(u)|]+C

Use the following diagram to change back to the variable x.

2x

√3

√3+4x2

u

From the diagram, tan(u) = 2x√3

and sec(u) =√

3+4x2√

3. Therefore,∫ (

4x2 +3)1/2

dx

=34

[2x√

3

√3+4x2√

3+ ln

∣∣∣∣∣√

3+4x2√

3+

2x√3

∣∣∣∣∣]+C

=34

[2x√

3

√3+4x2√

3+ ln

∣∣∣∣∣√

3+4x2√

3+

2x√3

∣∣∣∣∣]+C

=12

x√(3+4x2)+

34

ln∣∣∣√3+4x2 +2x

∣∣∣+C

Note that these examples involved something of the form(

a2 +(bx)2)

and the trigsubstitution bx = a tanu was the right one to use. This is the auspicious substitution whichoften simplifies these sorts of problems. However, there is a possibly better way to do thesekinds.

Example 8.5.4 Find∫ (

4x2 +3)1/2 dx another way.

Let 2x =√

3sinhu and so 2dx =√

3cosh(u)du. Then substituting in the integral leadsto ∫ √

3√

1+ sinh2 (u)

√3

2cosh(u)du =

32

∫cosh2 (u)du+C

=34

cosh(u)sinh(u)+34

u+C =34

√1+ sinh2 (u)sinh(u)+

34

u+C

=12

x√(3+4x2)+

34

sinh−1(

2x√3

)+C

This other way is often used by computer algebra systems. If you solve for sinh−1 x interms of ln, you get the same set of antiderivatives. The function sinh−1 is also written asarcsinh by analogy to the trig. functions also as asinh.