M408C: Hyperbolic Functions, Integration by

Parts, and Trigonometric IntegralsNovember 25, 2008

Let’s briefly review the definitions and properties of hyperbolic functions:

sinhx =ex − e−x

2coshx =

ex + e−x

2tanh x =

sinhx

coshx

csch x =1

sinhxsech x =

1coshx

coth x =coshx

sinhx

d

dx(sinhx) = coshx

d

dx(coshx) = sinhx

d

dx(tanh x) = sech2x

d

dx(csch x) = −csch x coth x

d

dx(sech x) = −sech x tanh x

d

dx(coth x) = −csch2x

Integration by parts is one of the most useful techniques of integration we will learn. It is agood technique to try when u-substitution does not work. The formula is simple:

where u = f(x), v = g(x), so du = f ′(x)dx and dv = g′(x)dx. Learning what to set u, v equal totakes some experience, but generally, one wants to set u to be a function that is easily differentiatedand dv to a function that is easily integrated. Thus it helps to think of setting u, dv instead of u, v.When there are bounds on the integral, make sure to carry over those bounds:∫ b

af(x) · g′(x)dx = f(x)g(x)

]b

a

−∫ b

ag(x) · f ′(x)dx.

Finally, we can use trigonometric identities to solve some rather thorny integrals of trigonometricfunctions. These integrals typically involve powers and products of trigonometric functions forwhich u-substitution is not helpful. The following formulas are generally the most helpful:

sin2 x + cos2 x = 1 sin2 x =12(1− cos 2x) cos2 x =

12(1 + cos 2x)

tan2 x = sec2 x− 1∫

tanx dx = ln |sec x|+ C

∫sec x dx = ln |sec x + tanx|+ C

1. 2.

1

3. (7.7.40) Differentiate y = 4

√1+tanh x1−tanh x .

Solution: We write 4

√1+tanh x1−tanh x as

(1+tanh x1−tanh x

) 14 and differentiate:

y′ =14

(1 + tanhx

1− tanh x

)− 34(

1 + tanhx

1− tanh x

)′

=14

(1 + tanhx

1− tanh x

)− 34(

sech2x(1− tanh x)− (1 + tanhx)(−sech2x)(1− tanh x)2

)=

14

(1− tanh x

1 + tanhx

) 34(

sech2x− sech2x tanh x + sech2x + sech2x tanh x

(1− tanh x)2

)=

14

1

(1 + tanhx)34

(1− tanh x)34

(1− tanh x)2(2 sech2x

)=

sech2x

2(1 + tanhx)34 (1− tanhx)

54

4. (8.1.18) Evaluate∫

e−θ cos 2θ dθ.Solution: We will be integrating by parts twice. Begin by setting u = e−θ, dv = cos 2θ, dθ, so

du = −e−θdθ, v = (sin 2θ)/2. We then get

I =∫

e−θ cos 2θ dθ =12e−θ sin 2θ −

∫12

sin 2θ(−e−θ)dθ =12e−θ sin 2θ +

12

∫e−θ sin 2θ dθ.

The last integral on the right resembles our original integral I: it is an exponential times a trigono-metric function. So we do integration by parts again: u = e−θ, dv = sin 2θ, dθ, so du = −e−θdθ,v = −(cos 2θ)/2. Then we get

I =12e−θ sin 2θ +

12

[−1

2e−θ cos 2θ −

∫ (−1

2cos 2θ

)(−e−θ)dθ

]=

12e−θ sin 2θ − 1

4e−θ cos 2θ − 1

4

∫e−θ cos 2θ dθ =

12e−θ sin 2θ − 1

4e−θ cos 2θ − 1

4I + C

Bringing the I/4 to the left, and then multiplying through, we get our solution:54I =

12e−θ sin 2θ − 1

4e−θ cos 2θ + C =⇒ I =

25e−θ sin 2θ − 1

5e−θ cos 2θ + C

5. (8.2.32) Evaluate I =∫

tan6(ay)dy for a ∈ R.Solution: Our strategy will be to convert tan2 x into sec2 x− 1 as often as we can.

I =∫

(tan4 ay)(tan2 ay)dy =∫

tan4 ay(sec2 ay − 1)dy =∫

tan4 ay sec2 ay dy −∫

tan4 ay dy

For the first integral on the right, we make the substitution u = tan ay, so du = a sec2 ay dy.Thesecond integral on the right is in the form of our original integral, so we again split it as before:

I =1a

∫u4du−

∫(tan2 ay)(tan2 ay)dy =

1a

u5

5−

∫tan2 ay(sec2 ay − 1)dy

=tan5 ay

5a−

∫tan2 ay sec2 ay dy +

∫tan2 ay dy

Setting u = tan ay, du = a sec2 ay dy, we are able to solve the middle integral

I =tan5 ay

5a− 1

a

∫u2du +

∫sec2 ay − 1 dy =

tan5 ay

5a− tan3 ay

3a+

∫sec2 ay dy −

∫1 dy

=tan5 ay

5a− tan3 ay

3a+

tan ay

a− y + C.

2