(1)(1) Using the concept of Riemann sum to evaluate (a) lim n→∞ 1 n n X k=1 sinkπ n = Z 1 0 sin πxdx

(1) Using the concept of Riemann sum to evaluate (a) lim

n→∞

1 n

n

X

k=1

sinkπ n =

Z 1 0

sin πxdx.

(b) lim

n→∞

n

X

k=1

n n²+ k² =

Z 1 0

1 1 + x²dx.

(c) lim

n→∞

1 n

n

X

k=1

r 1 + k

n = Z 1

0

√1 + xdx.

(d) lim

n→∞

e^1/n+ e^2/n+ · · · + e^n/n

n =

Z 1 0

e^xdx.

(e) lim

n→∞

n

X

k=1

ln ⁿ r

1 + k n =

Z 1 0

ln(1 + x)dx.

(f) lim

n→∞

√1 n

 1

√n + 1+ 1

√n +√

2+ · · · + 1

√n +√ n



= Z 1

0

dx 1 +√

x.

For a suitable choice of function f ∈ [0, 1], sums in the above limits can be written as Pn

n=1 1

nf ^k_n
which is a Riemann sum of f with respect to partition Pn= {k/n : 0 ≤ k ≤ n}.

The limit is thus given by the integralR1

0 f (x)dx.

(2) Evaluate (a)

Z

xⁿln xdx.

Solution: This can be solved by integration by parts.

Z

xⁿln xdx = Z

ln xd xⁿ⁺¹ n + 1



= xⁿ⁺¹ln x n + 1 −

Z xⁿ⁺¹ n + 1d ln x

=xⁿ⁺¹ln x n + 1 − 1

n + 1 Z

xⁿ⁺¹·dx

x = xⁿ⁺¹ln x n + 1 − 1

n + 1 Z

xⁿdx

=xⁿ⁺¹ln x

n + 1 − xⁿ⁺¹ (n + 1)² + C.

(b)

Z dx

sin²x + 4 cos²x.

Solution:

1

Let t = tan x. Then cos x = 1

√

1 + t² and sin x = ^√t

1+t² and dx = _1+t^dt2. The integral becomes

Z dx

sin²x + 4 cos²x =

Z 1

t

1+t² +_1+t⁴2

· dt 1 + t²

= Z dt

t + 4

= ln |t + 4| + C

= ln | tan x + 4| + C.

(c) Z

tan⁻¹1 xdx.

Solution: We use integration by parts:

Z

tan⁻¹ 1

xdx = x tan⁻¹ 1 x−

Z

x · d tan⁻¹ 1

x = x tan⁻¹ 1 x−

Z

x · 1

1 + (1/x)² · −dx x²

= x tan⁻¹ 1 x+

Z xdx 1 + x²

= x tan⁻¹ 1 x+1

2ln(1 + x²) + C.

(d) Z √

x − 1 x + 3 dx.

Solution:

Let u =√

x − 1. Then x = 1 + u² and dx = 2udu. Thus Z √

x − 1 x + 3 dx =

Z u

4 + u² · 2udu = 2 Z u²

u²+ 4du

= 2 Z 

1 − 4 u²+ 4

 du

= 2u − 4

Z 1

(u/2)²+ 1 du

2

= 2u − 4 tan⁻¹u 2 + C

= 2√

x − 1 − 4 tan⁻¹

√x − 1 2 + C.

(e)

Z x

(x + 1)(x + 2)(x + 3)dx.

Solution:

Use Partial fraction expansion.

(f)

Z 8x²+ 4x − 11 (x + 3)(x − 1)²dx.

Solution:

Use partial fraction expansion.

(g) Z

x²tan⁻¹xdx.

Solution:

We use integration by parts:

Z

x²tan⁻¹xdx = 1 3

Z

tan⁻¹xdx³

= x³

3 tan⁻¹x − 1 3

Z

x³· 1 1 + x²dx

= x³

3 tan⁻¹x − 1 3

Z x³ 1 + x²dx.

The use partial fraction expansion to evaluate

Z x³ 1 + x²dx.

(3) Evaluate (a)

Z π 0

√1 − sin xdx.

Write sin x = 2 sin^x₂cos^x₂. Then 1 − sin x =

cosx

2 − sinx 2

2

. Hence the integral becomes

Z π 0

| cosx

2 − sinx 2|dx.

(b) Z π/2

0

cos θ p2 − sin²θ

dθ.

Solution:

Let u = sin θ. The integral becomes Z 1

0

√ du

2 − u² = Z 1

0

1 q

1 − (u/√ 2)²

√du 2.

Let u =√

2 sin t. Then the integral becomes Z ^π₄

0

1dt = π 4. (c)

Z 1 0

x²+ 1 x⁴+ 1dx.

Solution:

Write

x⁴+ 1 = x⁴+ 2x²+ 1 − 2x²

= (x²+ 1) − (√ 2x)²

= (x²−√

2x + 1)(x²+√

2x + 1).

Then use partial fraction expansion.

(d) Z 1

0

√4

x 1 +√

xdx.

Solution:

(e) Z π/2

0

| cos 2x − sin x|dx.

Solution:

Find x in [0, π/2] such that cos 2x > sin x and cos 2x < sin x.

(f) Z 10

2

x + 1 x√

x − 1dx.

Solution:

(g) Z 1

0

x³e^x2 (x²+ 1)²dx.

Solution:

(h) Z 3

1

|x²− 4|dx.

Solution:

(i) Z 1

1/2

√ dt 2t − t².

Solution:

(j) Z 1

0

ln(1 + x)

ln(1 + x) + ln(2 − x)dx.

Solution:

(k) Z π/2

0

sin⁴x cos⁵xdx.

Solution:

(l) Z π

0

cosⁿx sin²(n + 1)xdx.

Solution:

(m) Z π/2

0

sin x sin 2x sin 3xdx.

Solution:

(n) Z 2π

0

dx

(2 + cos x)(3 + cos x).

Solution:

(4) Evaluate (a)

Z x · 3^x2 3^x2− 2dx.

Solution:

Let u = 3^x2− 2. Then du = 3^x2· ln 3 · 2xdx.

(b)

Z cos x

√4 − cos²xdx.

Solution:

Write 4 − cos²x = 3 + sin²x. The integral becomes Z cos x

p3 + sin²x dx.

Let u =√ 3 sin x.

(c)

Z x

√x²+ 2xdx.

Solution:

Write x²+ 2x = (x + 1)²− 1. Then the integral becomes

Z x

√x²+ 2xdx =

Z x

p(x + 1)²− 1dx =

Z x + 1

p(x + 1)²− 1dx −

Z 1

p(x + 1)²− 1dx.

For the first integral, we let u = (x + 1)². Then du = 2(x + 1)dx. Thus

Z x + 1

p(x + 1)²− 1dx =

Z du

2√

u − 1 =√

u − 1 + C =p

x²+ 2x + C.

(d) Z

sech⁻¹(x)dx.

Solution:

(e)

Z u

u⁶− 8du.

Solution:

(f) Z

sin⁶xdx.

Solution:

(g)

Z 1

1 + e^xdx.

Solution:

(h) Z

sin(ln x)dx.

Solution:

(5) Evaluate (a)

Z 4 1

sec⁻¹√ xdx.

Solution:

(b) Z 2

0

√x

√x +√

2 − xdx.

Solution:

(c) Z π

−π

(x²p³

x³+ 1 + | cos x|)dx.

Solution:

(d) Z 1

0

tan⁻¹x 1 + x dx.

Solution:

(e) Z 2

0

x⁵e^−x3dx.

Solution:

Write

x⁵e^−x3 =



−x³ 3



−3x²e^−x3 . Thus we may use integration by parts to solve the problem:

Z 2 0

x⁵e^−x3dx = Z 2

0



−x³ 3

 de^−x3

= −x³

3 e^−x3|²₀− Z 2

0

−x²e^−x3dx

= −x³

3 e^−x3|²₀−1 3e^x3|²₀. (f)

Z 1/2 0

√ dx

x²+ 4x + 5.

Solution:

write x²+ 4x + 5 = (x + 2)²+ 1. Taking u = x + 2, the integral becomes Z ³₂

2

√ du u²+ 1.

(g) Z

√π/2

0

d

du[sin u cos u²]du.

Solution:

By the fundamental theorem of calculus:

Z

√π/2

0

d

du[sin u cos u²]du = sin u cos u²|

√π/2 0 .

(h) Z π/2

0

sin x

cos²x + 3 cos x + 2dx.

Solution:

Let t = cos x. Then the integral becomes Z 1

0

dt t²+ 3t + 2. Then use partial fraction expansion.

(i) Z π/4

0

dθ

(tan²θ + 4 tan θ + 3) cos²θ. Solution:

Write the integral as Z π/4

0

dθ

(tan²θ + 4 tan θ + 3) cos²θ = Z π/4

0

sec²θdθ (tan²θ + 4 tan θ + 3). Let u = tan θ. Then the integral becomes

Z 1 0

du u²+ 4u + 3. Then use partial fraction expansion.

(j) Z 7π/3

π/3

sin³xdx.

Solution:

Write sin³x = sin x · sin²x = sin x(1 − cos²x). Let u = cos x.

(6) Evaluate I1=

Z sin x

cos x + sin xdx and I2=

Z cos x cos x + sin xdx.

Hint: evaluate both I1+ I2and I1− I2.

We know

I1+ I2= Z

1dx = x + C I1− I2=

Z sin x − cos x sin x + cos xdx.

Let u = sin x + cos x. We have du = (cos x − sin x)dx. Thus I₁− I2= −

Z du

u = − ln |u| + C = − ln | sin x + cos x| + C.

This implies that I1=1 2x − 1

2ln | sin x + cos x| + C and I2=1 2x +1

2ln | sin x + cos x| + C

(7) Suppose a < b. Evaluate Z b

a

p(x − a)(b − x)dx.

Solution:

Let x = (b − a) sin²t + a. Then dx = 2(b − a) sin t cos tdt and x − a = (b − a) sin²t, b − x = (b − a) cos²t.

Thus Z

p(x − a)(b − x)dx = Z

(b − a) sin t cos t · 2(b − a) sin t cos tdt

= 1

2(b − a)² Z

sin²2tdt

= 1

2(b − a)²

Z 1 − cos 2t

2 dt

= 1

4(b − a)²

 t −1

2sin 2t

 + C

= 1

4(b − a)² sin⁻¹r x − a

b − a −p(x − a)(b − x) b − a

! + C.

(8) Suppose a > 0 ac − b²> 0. Show that Z π/2

−π/2

dx

a cos²x + 2b cos x sin x + c sin²x = π

√ac − b². Solution:

Let I be the integral and t = tan x. Then dt

1 + t² = dx and sin x = t

√1 + t², cos x = 1

√1 + t². Thus the integral becomes

Z ∞

−∞

dt a + 2bt + ct². Completing the square of the enumerator, we obtain

a + 2bt + ct²= c

 t +b

c

²

+ac − b² c

Then Z ∞

−∞

dt

a + 2bt + ct² = Z ∞

−∞

cdt

(ct + b)²+ ac − b². Let u = ct + b. Then du = cdt. Thus

I = Z ∞

−∞

du u²+ ac − b². Let v = u

√

ac − b². Then dv = du

√ac − b²

. Hence

I = 1

√ ac − b²

Z ∞

−∞

dv v²+ 1. We know

Z ∞

−∞

dv

v²+ 1 = π. Thus I = π

√ac − b².