18.01 Practice Exam 5

1. (5 points each) Evaluate:

(a) tan^{−1} 1;

Solution: tan^{−1} 1 = \frac{π}{4} .

(b) cosh(ln 2).

Solution: cosh(ln 2) = \frac{1}{2}(e^{ln 2} + e^{− ln 2}) = \frac{1}{2}(2 + 2^{−1}) = \frac{5}{4}.

2. (5 points each)

(a) For what values of of x is f(x) = sin^{−1}(e^{x}) defined?

Solution: We need −1 ≤ e^{x} ≤ 1. Since e^{x} > 0 for any x, we always have −1 ≤ e^{x}, but e^{x} ≤ 1 is only satisfied for x ≤ 0. So f(x) is defined for x ≤ 0.

(b) Find f′(x).

Solution:

f′(x) = \frac{d}{dx} (sin^{−1}(e^{x}))
= \frac{1}{\sqrt{1 − (e^{x})^{2}}} · \frac{d}{dx}(e^{x}) = \frac{e^{x}}{\sqrt{1 − e^{2x}}} .

3. (10 points each) Evaluate the following definite integrals:

(a) ∫_{0}^{π/4} cos^{5} x dx;

Solution: Substitute u = sin x dx, du = cos x dx, and use cos^{2} x = 1 − sin^{2} x. This gives

∫_{0}^{π/4} cos^{5} x dx = ∫_{0}^{π/4}(cos^{2} x)^{2} cos x dx
= ∫_{0}^{π/4}(1 − sin^{2} x)^{2} cos x dx

= ∫_{0}^{1/\sqrt{2}}(1 − u^{2})^{2} du

= ∫_{0}^{1/\sqrt{2}}(1 − 2u^{2} + u^{4}) du

= (u − \frac{2}{3}u^{3} + \frac{1}{5}u^{5})]_{0}^{1/\sqrt{2}}

= \frac{1}{\sqrt{2}} − \frac{2}{3 · 2\sqrt{2}} + \frac{1}{5 · 4\sqrt{2}}
= \frac{1}{\sqrt{2}} (1 − \frac{1}{3} + \frac{1}{20})
= \frac{43}{60\sqrt{2}}.

(b) ∫_{1}^{e} \sqrt{x} ln x dx. Solution: Let u = ln x, dv = x^{1/2} dx, so du = \frac{1}{x} dx and v = \frac{2}{3}x^{3/2} and the indefinite integral is:

∫ \sqrt{x} ln x dx = \frac{2}{3}x^{3/2} ln x − ∫ \frac{2}{3}x^{1/2} dx = \frac{2}{3}x^{3/2} ln x − \frac{4}{9}x^{3/2} + C.

Therefore

∫_{1}^{e} \sqrt{x} ln x dx = (\frac{2}{3}x^{3/2} ln x − \frac{4}{9}x^{3/2})]_{1}^{e}
= \frac{2}{3}e^{3/2} − \frac{4}{9}e^{3/2} + \frac{4}{9}
= \frac{2}{9}e^{3/2} + \frac{4}{9}

4. (10 points each) Find the following indefinite integrals:

(a) ∫ x cosh(x^{2}) dx;

Solution: Substitute u = x^{2}, du = 2x dx, so

\frac{1}{2} ∫ cosh(x^{2}) · 2 dx = \frac{1}{2} ∫ cosh u du
= \frac{1}{2} ∫ sinh u + C = \frac{1}{2} ∫ sinh(x^{2}) + C.

(b) ∫ \frac{1}{\sqrt{2x − x^{2}}} dx;

Solution: Complete the square by writing

2x − x^{2} = 1 − 1 + 2x − x^{2} = 1 − (x − 1)^{2}.

Now substituting u = x − 1, du = dx gives

∫ \frac{1}{\sqrt{1 − (x − 1)^{2}}} dx = ∫ \frac{1}{\sqrt{1 − u^{2}}} du
= sin^{−1} u + C = sin^{−1}(x − 1) + C.

(c) ∫ \frac{1}{x^{3} − x^{2}} dx;

Solution: Use partial fractions. Since x^{3} − x^{2} = x^{2}(x − 1), set

\frac{1}{x^{3} − x^{2}} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x − 1},

and find A, B and C by solving:

1 = Ax(x − 1) + B(x − 1) + Cx^{2} = (A + C)x^{2} + (B − A)x − B. Equating coefficients gives A + C = 0, B − A = 0 and −B = 1, so B = −1, A = −1 and C = 1 and

∫ \frac{1}{x^{3} − x^{2}} dx = − ∫ \frac{dx}{x} − ∫ \frac{dx}{x^{2}} + ∫ \frac{dx}{x − 1}
= − ln |x| + \frac{1}{x} + ln |x − 1| + C.

(d) ∫ \frac{x^{2}}{\sqrt{a^{2} − x^{2}}} dx;

Solution: Substitute x = a sin θ, so θ = sin^{−1} \frac{x}{a}, dx = a cos θ dθ and \sqrt{a^{2} − x^{2}} = a cos θ. This gives

∫ \frac{x^{2}}{\sqrt{a^{2} − x^{2}}} dx = ∫ \frac{a^{2} sin^{2} θ}{a cos θ} a cos θ dθ
= a^{2} ∫ sin^{2} θ dθ
= \frac{a^{2}}{2} ∫ (1 − cos(2θ)) dθ
= \frac{a^{2}}{2} (θ − \frac{1}{2} sin(2θ))
= \frac{a^{2}}{2} (θ − sin θ cos θ)
= \frac{a^{2}}{2} (sin^{−1} \frac{x}{a} − \frac{x}{a^{2}} \sqrt{a^{2} − x^{2}})
= \frac{a^{2}}{2} sin^{−1} \frac{x}{a} − \frac{x}{2}\sqrt{a^{2} − x^{2}}

5. (10 points) Use the reduction formula

∫ sec^{n} u du = \frac{1}{n − 1} sec^{n−2} u tan u + \frac{n − 2}{n − 1} ∫ sec^{n−2} u du

to find ∫ \frac{x^{4}}{\sqrt{x^{2} − 1}} dx.

Solution: Substitute x = sec θ, so θ = sec^{−1} x, dx = sec θ tan θ dθ and \sqrt{x^{2} − 1} = tan θ. This gives

∫ \frac{x^{4}}{\sqrt{x^{2} − 1}} dx = ∫ \frac{sec^{4} θ}{tan θ} tan θ dθ
= ∫ sec^{5} θ dθ
= \frac{1}{4} sec^{3} θ tan θ + \frac{3}{4} ∫ sec^{3} θ dθ
= \frac{1}{4} sec^{3} θ tan θ + \frac{3}{8} sec θ tan θ + \frac{3}{8} ∫ sec θ dθ
= \frac{1}{4} sec^{3} θ tan θ + \frac{3}{8} sec θ tan θ + \frac{3}{8} ln | sec θ + tan θ| + C
= \frac{1}{4}x^{3}\sqrt{x^{2} − 1} + \frac{3}{8}x\sqrt{x^{2} − 1} + \frac{3}{8} ln |x + \sqrt{x^{2} − 1}| + C

6. (5 points each)

(a) State the formula for integration by parts (assume u and v are differentiable functions of x).

Solution:

∫ u dv = uv − ∫ v du.

(b) Derive the formula from a basic differentiation rule.

Solution: The product rule states that \frac{d}{dx}(uv) = u\frac{dv}{dx} + v \frac{du}{dx}, so

u\frac{dv}{dx} = \frac{d}{dx}(uv) − v \frac{du}{dx}.

Integrating both sides of the equation with respect to x gives

∫ u\frac{dv}{dx} dx = ∫ \frac{d}{dx}(uv) dx − ∫ v \frac{du}{dx} dx,

which is the formula for integration by parts.