Name:

UIN:

MATH-172-501

Answers to Second Midterm

Javier Elizondo

I. Find the integral ∫sin2 x cos2 x dx

Sol. sin2 x cos2 x = (sin x cosx)2 =(

12

sin(2x))2

= 14

sin2(2x) = 18

(1− cos(4x)) Now

the integral becomes∫

18

(1− cos(4x)) dx which is very easy to compute.

II. Find the integral ∫dx

x√x2 + 3

.

Sol. x =√

3 tan θ, where −π2< θ < π

2, and dx =

√3 sec2 θ dθ. We also now that√

3 tan2 +3 =√

3√

sec2 x =√

3 secx. Therefore the integral becomes1√3

∫sec θ dθtan θ

= 1√3

∫csc θ dθ. We use the formula written in the exam, and with the

appropriate right rectangle we return to the variable x.

III. Find ∫x2

(x− 3)(x+ 2)2dx

Sol. x2

(x−3)(x+2)2= A

x−3+ B

x+2+ C

(x+2)2=⇒ x2 = A(x+ 2)2 +B(x− 3)(x+ 2) +C(x− 3).

If x = 3 we get A = 9/25, if x = −2 we get C = −4/5. If we compare the coefficient

of x2 in the right hand side and left hand side polynomials of the equality we see that

1 = A+B ⇒ B = 16/25. Then the integral is equal to925

ln |x− 3|+ 1625

ln |x+ 2|+ 45(x+2)

+ C.

IV. Compute the improper integral ∫ ∞0

xe−x dx

Sol. Using parts with du = e−xdx and v = x we obtain∫∞0xe−x dx = limt→∞ (−xe−x − e−x)t0 = limt→∞ (1− (t+ 1)e−t) = 1 − limt→∞

t+1et by

L’Hpital this last limit is equal to 1− limt→∞1et = 1− 0 = 1. Therefore the integral is

convergent and is equal to 1.

V. A tank contains 100 L of pure water. Brine that contains 0.1 kg of salt per liter enters

the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from

the tank at the same rate. How much salt is in the tank after t minutes?

Sol. The initial condition is y(0) = 0 where as usual y(t) is the amount of salt at time

t. Now, dydt

= rate in - rate out = (0.1 × 10)−(y

100× 10

)= 1− y

10= 10−y

10⇒∫

dy10−y =

∫110dt ⇒ − ln |10 − y| = 1

10t + C ⇒ 10 − y = Ae−t/10. The initial condition

implies A = 10, therefore y = 10(1− e−t/10

).

VI. Solve the initial-value problem

x2 dy

dx+ 2xy = cosx, y(π) = 0

Sol. y′ = 2xy = cosx

x2 , and I(x) = eR

(2/x)dx = x2. Multiplying the differential equation

by I(x) we obtain x2y′ + 2xy = cosx ⇒ (x2y)′ = cosx ⇒ y = 1x2

(∫cosxdx+ C

)=

1x2 (sinx+ C) =⇒ y = sinx

x2 where the last equality follows from the fact that the initial

condition implies that C = 0.