Let

In the denominator, we can write

cos²x + 3 cos x + 2 = (cos x + 1)(cos x + 2)

Put cos x = t

⇒ –sin(x)dx = dt (Differentiating both sides)

⇒ sin(x)dx = –dt

When x = 0, t = cos 0 = 1

So, the new limits are 1 and 0.

Substituting this in the original integral,

We can write,

Using this, we have

Recall

⇒ I = – [2(ln|0+2| – ln|1+2|) – (ln|0+1| – ln|1+1|)]

⇒ I = – [2(ln 2 – ln 3) – (ln 1 – ln 2)]

⇒ I = – (2 ln 2 – 2 ln 3 – 0 + ln 2)

⇒ I = – (3 ln 2 – 2 ln 3)

⇒ I = 2 ln 3 – 3 ln 2

⇒ I = ln 9 – ln 8 = ln