We had two problems in class on integration that we write up here.
1.
∫ 𝑥 sin 𝑥 cos 𝑥 𝑑𝑥
As we started in class, we let 𝑑𝑣 = sin 𝑥 cos 𝑥 𝑑𝑥, 𝑢 = 𝑥. Then 𝑣 = sin2 𝑥 and 𝑑𝑢 = 𝑑𝑥. By te
integration by parts formula, we have that the original integral is equal to
𝑥 sin2 𝑥 − ∫ sin2 𝑥 𝑑𝑥.
To solve the remaining integral, we use the half angle formula, sin2 𝑥 =
Consequently,
1−cos(2𝑥)
.
2
1
1
1
1
∫ sin2 𝑥 𝑑𝑥 = ∫ ( ) 𝑑𝑥 − ∫ cos(2𝑥) 𝑑𝑥 = 𝑥 − sin(2𝑥) + 𝐶,
2
2
2
4
Where in the first integral we used the power rule in reverse and in the second we used
the substitution 𝑤 = 2𝑥.
2. ∫ sec 3 𝑥 𝑑𝑥
We approach this by integration by parts using 𝑢 = sec(𝑥) , 𝑑𝑣 = sec 2 𝑥 𝑑𝑥, and so 𝑑𝑢 =
sec(𝑥) tan(𝑥) , 𝑣 = tan(𝑥). By the integration by parts formula we have
∫ sec 3 𝑥 𝑑𝑥 = sec(𝑥) tan(𝑥) − ∫ sec(𝑥) tan2 𝑥 𝑑𝑥.
For the remaining integral, we use the trig identity tan2 𝑥 = sec 2 𝑥 − 1, transforming the
integral:
∫ sec(𝑥) tan2 (𝑥) 𝑑𝑥 = ∫ sec 3(𝑥) 𝑑𝑥 − ∫ sec(𝑥) 𝑑𝑥.
The first integral here is our original integral so we leave it aside for a moment. We
can evaluate the second integral using a sneaky trick:
∫ sec(𝑥) 𝑑𝑥 = ∫
sec(𝑥) (sec(𝑥) + tan(𝑥))
𝑑𝑥 = ∫(sec 2 (𝑥) + sec(𝑥) tan(𝑥))/(sec(𝑥) + tan(𝑥)) 𝑑𝑥
sec(𝑥) + tan(𝑥)
We now use a substitution: 𝑤 = sec(𝑥) + tan(𝑥) , 𝑑𝑤 = sec(𝑥) tan(𝑥) + sec 2(𝑥). So,
∫(sec 2(𝑥) + sec(𝑥) tan(𝑥))/(sec(𝑥) + tan(𝑥)) 𝑑𝑥 = ∫ 𝑤 −1 𝑑𝑤 = ln|𝑤| + 𝐶
= ln | sec(𝑥) + tan(𝑥)| + 𝐶.
Putting this all together,
∫ sec 3 𝑥 𝑑𝑥 = sec(𝑥) tan(𝑥) − ∫ sec 3(𝑥)𝑑𝑥 + ln|sec(𝑥) + tan(𝑥)| + 𝐶.
Solving for the original integral yields,
1
∫ sec 3 𝑥 𝑑𝑥 = (sec(𝑥) tan(𝑥) + ln|sec(𝑥) + tan(𝑥)|) + 𝐶.
2
Note this is a tricky problem. If we were to put such a problem on an exam, there
would be lots of hints.