Page 166 - Calculus Demystified

CHAPTER 6
SOLUTION Transcendental Functions 153

4 1
dx = 4 dx = 4ln |x + 1|+ C
x + 1 x + 1

1 1
dx = ln |− 2 + 3x|+ C.
−2 + 3x 3
You Try It: Calculate the integral

cos x
dx.
2 + sin x
You Try It: Calculate the integral
e 1

dx.
3/2
1 x ·[ln x]
EXAMPLE 6.5
Evaluate the integral

cos x
dx.
3sin x − 4
SOLUTION
For clarity we set ϕ(x) = 3 sin x − 4, ϕ (x) = 3(cos x). The integral then

has the form

1 ϕ (x) 1
dx = ln |ϕ(x)|+ C.
3 ϕ(x) 3
Resubstituting the expression for ϕ(x) yields that

cos x 1
dx = ln |3 sin x − 4|+ C.
3 sin x − 4 3
2
x
You Try It: Evaluate dx.
1 − x 3
EXAMPLE 6.6
Calculate

cot xdx.

SOLUTION
We rewrite the integral as

cos x
dx.
sin x