Integrals and Trig Functions

Question

1. Deduce that ∫1/cos(x) dx is equal to both 1/2 ln(|1+sin(x)/1-sin(x)|) + C and ln(|sec(x) +tan(x)|) +C
2. Deduce that ∫1/cos^2(x) is equal to tan(x) + C

Answer 1

Hi Sydney!

I'll show you how to do the first part of the first one.

1 cos x

Rewrite ∫ ( ----- ) dx as ∫ ( ----------) dx and substitute cos² x with 1-sin² x (pythagorean identity)

cos x cos² x

so you have

cos x

∫ ( ------- ) dx

1-sin² x

cos x

∫ ( --------------------) dx Factor -- difference of two squares A² - B²

(1+sinx)(1-sinx)

Now, can do partial fraction expansion.

cos x A cosx B cosx

----------------------- = ----------- + ----------

(1+sinx)(1-sinx) (1+sinx) (1-sinx)

So then, just looking at the numerators:

cos x = (Acosx)(1-sinx) + (Bcosx)(1+sinx)

cos x = Acosx - Acosxsinx + Bcosx + Bcosxsinx

There is no cosxsinx term, so -Acosxsinx + Bcosxsinx = 0 or -A + B = 0

Acosx + Bcosx = 1cosx, so A + B = 1

Adding these two equations together gives 2B = 1 or B = 1/2.

It follows that A = 1/2 also.

So, our integral becomes

1 cos x 1 cosx

∫ [ -- ( -------- ) + -- ( ------- ) ] dx

2 1+sinx 2 1-sinx

1 cosx 1 cosx

-- ∫ ------- dx + --- ∫ -------- dx break up into separate integrals

2 1+sinx 2 1-sinx

let u₁ = 1+sinx let u₂ = 1-sinx

then du₁ = cosxdx then du₂ = -cosxdx or -du₂ = cosxdx

1 du₁ 1 -du₂

-- ∫ ---- + --- ∫ ----

2 u₁ 2 u₂

1 1

-- ln |u₁| - -- ln |u₂| + C

2 2

1 1

-- ln |1+sinx| - -- ln |1-sinx| + C Substitute in for the u's

2 2

1

-- [ ln|1+sinx| - ln|1-sinx| ] + C Factor out a 1/2

2

1 1+sinx

-- [ ln | ------- | + C Properties of logarithms: Log a - Log b = Log (a/b)

2 1-sinx

Hint for the second part of #1. Rewrite 1/cosx as secx and then multiply by

secx + tanx

----------------

secx + tanx

then let u = secx + tanx

Answer 2

the way the question is worded it could be answered

d(ln(sec(x)+tan(x))+C)/dx = 1/cos(x)

which is not hard to show

so by the fundamental theorem of calculus

integral of 1/cos(x) is ln(sec(x)+tan(x))+C

then it could be shown sec(x)+tan(x)=[(1+sin(x)/(1-sin(x))]^(1/2)

(1+sin(x))²/cos²(x)=(1+sin(x))/(1-sin(x))

note that

cos²(x)=(1+sin(x))(1-sin(x))

now have

(1+sin(x))²=(1+sin(x))²

same with integral of 1/cos²(x) equaling tan(x)+C

d(tan(x)+C)/dx=1/(cos²(x)

by the fundamental theorem of calculus

tan(x)+C=the integral of 1/cos²(x)