192 CHAPTER 8. METHODS FOR FINDING ANTIDERIVATIVES

Let u = 3x2so that du

dx = 2x ln(3)3x2and du

2ln(3) = x3x2dx. Thus

∫x3x2

dx =1

2ln(3)

∫du =

12ln(3)

[u+C] =1

2ln(3)3x2

+

(1

2ln(3)

)C

Since the constant is an arbitrary constant, this is written as 12ln(3)3x2

+C.

Example 8.1.5 Find∫

cos2 (x) dx

Recall that cos(2x) = cos2 (x)− sin2 (x) and 1 = cos2 (x)+ sin2 (x). Then subtractingand solving for cos2 (x),

cos2 (x) =1+ cos(2x)

2.

Therefore, ∫cos2 (x) dx =

∫ 1+ cos(2x)2

dx

Now letting u = 2x, du = 2dx and so∫cos2 (x) dx =

∫ 1+ cos(u)4

du =14

u+14

sinu+C =14(2x+ sin(2x))+C.

Also∫

sin2 (x) dx =− 12 cosxsinx+ 1

2 x+C which is left as an exercise. This trick involvinga trig. identity is almost the only way to do these.

Example 8.1.6 Find∫

tan(x) dx

Let u = cosx so that du = −sin(x)dx. Then writing the antiderivative in terms of u,this becomes

∫ −1u du. At this point, recall that (ln |u|)′ = 1/u. Thus this antiderivative is

− ln |u|+C = ln∣∣u−1

∣∣+C and so∫

tan(x) dx = ln |secx|+C.This illustrates a general procedure.

Procedure 8.1.7 ∫ f ′(x)f (x) dx = ln | f (x)|+C.

This follows from the chain rule and the derivative of x → ln |x|.

Example 8.1.8 Find∫

sec(x)dx.

This is usually done by a trick. You write as∫ sec(x)(sec(x)+tan(x))

(sec(x)+tan(x)) dx and note that thenumerator of the integrand is the derivative of the denominator. Thus∫

sec(x)dx = ln |sec(x)+ tan(x)|+C.

Example 8.1.9 Find∫

csc(x)dx.

This is done like the antiderivatives for the secant. ddx csc(x) = −csc(x)cot(x) and

ddx cot(x) =−csc2 (x) . Write the integral as

−∫ −csc(x)(cot(x)+ csc(x))

(cot(x)+ csc(x))dx =− ln |cot(x)+ csc(x)|+C.