The 2014 Integration Bee—Solutions and comments

Mike Hirschhorn

Qualifying Round

1.∫

3x2(x3 − 1)4 dx =

∫

u4 du =1

5u5 + C =

1

5(x3 − 1)5 + C.

2.∫

√

5

√

2

x√x2 − 1

dx =[

√

x2 − 1]

√

5

√

2= 2− 1 = 1.

3.∫

sin2(x

2

)

dx =

∫

1

2− 1

2cos x dx =

1

2x− 1

2sinx+ C.

4.∫

1

a2 + x2dx =

1

atan−1

(x

a

)

+C.

5.∫

3x− 2

2x+ 1dx =

∫

3

2− 7

2

1

2x+ 1dx =

3

2x− 7

4log |2x+ 1|+C.

6.∫ π

2

0

1

1 + sin θdθ =

∫ 1

0

1

1 +2t

1 + t2

· 2

1 + t2dt

=

∫ 1

0

2

(1 + t)2dt =

[

− 2

1 + t

]1

0= −1 + 2 = 1.

7.∫

x2014 dx =x2015

2015+ C.

8.∫

1

x2 − 4dx =

∫

1

4

(

1

x− 2− 1

x+ 2

)

dx =1

4log∣

∣

∣

x− 2

x+ 2

∣

∣

∣+ C.

Typeset by AMS-TEX

1

2

9.∫

ex(cos x+ sinx) dx = Aex cos x+Bex sinx+ C

for some A, B. Differentiating with respect to x gives

ex(cos x+ sinx) = (A+B)ex cos x+ (−A+B)ex sinx.

Comparing coefficients gives A+B = 1, −A+B = 1, A = 0, B = 1, so

∫

ex(cos x+ sinx) dx = ex sinx+ C.

Alternatively, you can do two integrations by parts.

10.∫

1

x log xdx =

∫

1

udu = log |u|+ C = log | log x|+C.

11.∫

πx dx =

∫

ex log π dx =1

log πex log π + C =

1

log ππx + C.

12.∫

1

3x2dx =

1

3· − 1

x+ C = − 1

3x+ C.

13.∫ 2

1

2x√

x2 − 1 dx =[2

3(x2 − 1)

3

2

]2

1=

2

3(3

3

2 − 0) = 2√3.

14.∫

x2 + 3

x2 − 3dz =

x2 + 3

x2 − 3z + C.

(We must assume x is a constant function of z.)

15.∫

x5 log |x| dx =x6

6log |x| −

∫

x5

6dx =

x6

6log |x| − x6

36+ C

by integration by parts with u = log |x|, dv = x5 dx.

3

16.

Let

∫ π

0

x sinx

1 + cos2 xdx = I. If we make the substitution x = π − u, we find

I =

∫ π

0

(π − u) sinu

1 + cos2 udu.

It follows that

2I =

∫

π sinx

1 + cos2 xdx = π

∫ 1

−1

1

1 + u2du = π

[

tan−1 u]1

−1= π

(π

4−(

−π

4

))

=π2

2,

so∫ π

0

x sinx

1 + cos2 xdx = I =

π2

4.

17.

∫ √1 + sin 2x dx =

∫

√

(cos x+ sinx)2 dx =

∫

cos x+sinx dx = − cos x+sinx+C.

(Not exactly legitimate, I’m afraid. It is not always the case that√v2 = v.) This

is one reason why definite integrals make better quiz questions than indefinite

integrals.

18.

∫ π

2

0

sinx

1 + sin2 xdx =

∫ π

2

0

sinx

2− cos2 xdx =

∫ 1

0

1

2− u2du =

∫ 1

0

1

2√2

(

1√2− u

+1√2 + u

)

du

=1

2√2

[

log∣

∣

∣

√2 + u√2− u

∣

∣

∣

]1

0=

1

2√2log

√2 + 1√2− 1

=1√2log(1 +

√2).

19.

∫ 9π

10

π

7

√cosecx− sinx dx =

∫ 9π

10

π

7

√

1

sinx− sinx dx =

∫ 9π

10

π

7

√

1− sin2 x

sinxdx

=

∫ 9π

10

π

7

cos x√sinx

dx =[

2√sinx

]9π

10

π

7

= 2

(

√

sin9π

10−√

sinπ

7

)

.

4

20.∫

cotx dx =

∫

cos x

sinxdx = log | sinx|+ C.

21.∫ 5

−5

|π − x| dx =

∫ π

−5

π − x dx+

∫ 5

π

x− π dx =[

πx− 1

2x2]π

−5+[1

2x2 − πx

]5

π

= (π2 − 1

2π2)− (−5π − 25

2) + (

25

2− 5π)− (

1

2π2 − π2) = π2 + 25.

Alternatively, from a graph,

∫ 5

−5

|pi− x| dx =1

2(π + 5)2 +

1

2(5− π)2 = π2 + 25.

22.∫

e5 log x dx =

∫

x5 dx =x6

6+ C.

23.∫

eex+x dx =

∫

eex · ex dx =

∫

eu du = eu + C = eex

+ C.

24.∫

sec x tanx dx = sec x+ C.

(This is/used to be bookwork.)

25.∫

log x

xdx =

∫

1

udu = log |u|+ C = log | log x|+ C.

26.∫

1√4− x2

dx = sin−1(x

2

)

+ C.

(This certainly is bookwork!

27.

∫

1

cos x+ 1dx =

∫

1

1− t2

1 + t2+ 1

· 2

1 + t2dt =

∫

1 dt = t+ C = tan−1(x

2

)

+ C.

5

28.∫

cosec2θ dθ = − cot θ + C.

This used to be bookwork once. Differentiate the right side to check!

29.

∫

x3e3x2

dx =

∫

1

18(3x2)(6x)e3x

2

dx =

∫

1

18ueu du =

1

18(ueu − eu) + C

=1

18

(

3x2e3x2 − e3x

2)

+ C =1

18e3x

2

(3x2 − 1) + C.

30.

∫

tan−1 x dx =

∫

u sec2 u du = u tanu−∫

tanu du = u tan u+ log | cos u|+ C

= u tan u− log | sec u|+ C = u tanu− 1

2log(1 + tan2 u) + C

= x tan−1 x− 1

2log(1 + x2) + C.

Alternatively, do an integration by parts with u = tan−1 x, dv = dx.

6

Group Stage

1.

sin(A+B) = sinA cosB + cosA sinB,

so∫ π

4

π

8

sin 5x cos 3x dx =

∫ π

4

π

8

1

2(sin 8x+ sin 2x) dx =

[1

2

(

−cos 8x

8− cos 2x

2

)

]π

4

π

8

= − 1

16

[

cos 8x]

π

4

π

8

− 1

4

[

cos 2x]

π

4

π

8

= − 1

16(1− (−1))− 1

4

(

0− 1√2

)

= −1

8+

1

4√2

=

√2− 1

8.

2.∫ 16π

0

sin2 x cos2 x dx =

∫ 16π

0

1

4sin2 2x dx =

∫ 16π

0

1

8(1− cos 4x) dx =

[1

8x− 1

32sin 4x

]16π

0

= 2π.

3.∫ 5

−5

|5− x| dx =

∫ 5

−5

5− x dx =[

− 1

2(5− x)2

]5

−5=

1

2102 = 50.

Look at the graph!

4.∫ 3

2

1

2

2− 2x√2x− x2

dx =

∫ 3

2

1

2

2(1 − x)√

1− (1 − x)2dx =

∫

−1

2

1

2

2u√1− u2

du = 0,

because the integrand is odd and the interval of integration is balanced about the

origin.

5.

∫ 1

5

0

2√4− 25x2

dx =2

5

∫ 1

5

0

1√

(

25

)2 − x2

dx =2

5

[

sin−1 5x

2

]1

5

0=

2

5· π6=

π

15.

6.∫

cos x

4− sin2 xdx =

∫ 1

0

1

4− u2du =

∫ 1

0

1

4

(

1

2 + u+

1

2− u

)

du =1

4

[

log∣

∣

∣

2 + u

2− u

∣

∣

∣

]1

0

=1

4log 3.

7

7.∫ 1

0

esin2 xecos

2 x dx =

∫ 1

0

e dx = e.

8.

∫ 1

0

1

x2 − 2x+ 5dx =

∫ 1

0

1

(x− 1)2 + 22dx =

[1

2tan−1 x− 1

2

]1

0= −1

2tan−1

(

−1

2

)

=1

2tan−1

(

1

2

)

.

9.

∫ 2

1

x3ex2

dx =

∫ 2

1

1

2(2x)x2ex

2

dx =

∫ 4

1

1

2ueu du =

[1

2(ueu − eu)

]4

1=

3

2e4.

10.

∫ π

4

0

sinx sin 2x dx =

∫ π

4

0

1

2(cos x− cos 3x) dx =

[ sinx

2− sin 3x

6

]π

4

0=

1

2√2− 1

6√2

=2

6√2=

1

3√2.

11.∫ 1

2

1

3

e1

x

x3dx =

∫ 3

2

ueu du =[

ueu − eu]3

2= 2e3 − e2.

12.

∫ 5

2

cos(sinx) cos x dx =

∫ sin 5

sin 2

cosu du =[

sinu]sin 5

sin 2= sin(sin 5)− sin(sin 2).

13.

e

∫

(x2 − 3x+ 2)−1 dx= e

∫

1

x− 1− 1

x− 2dx

= elog∣

∣

∣

x− 1

x− 2

∣

∣

∣+C

= eC∣

∣

∣

x− 1

x− 2

∣

∣

∣.

14.∫ 1

0

ex tan ex dx =

∫ e

1

tanu du =[

log | sec u|]e

1= log

∣

∣

∣

sec e

sec 1

∣

∣

∣.

8

That’s what a blind calculation gives, but it is not correct, would you believe!

Sketch the graph of the integrand, and see what goes wrong! It makes you wonder

about some of the other integrals!

15.∫ 1

0

xe−x dx =[

− xe−x − e−x]1

0= 1− 2e−1.

The indefinite integral can be guessed, or you can do an integration by parts.

16.

∫

(ex + 1)20ex dx =

∫ e

1

(u+ 1)20 du =[ 1

21(u+ 1)21

]2

1=

1

21

(

(e+ 1)21 − 221)

.

17.∫ π

2

π

3

sinx cos2 x dx =

∫ 1

2

0

u2 du =[1

3u3]

1

2

0=

1

24.

18.∫ 1

0

1

1 + x2dx =

[

tan−1 x]1

0= tan−1 1 =

π

4.

19.

∫ π

4

π

12

tan x+ cot x dx =[

log | sec x|+ log | sinx|]

π

4

π

12

=[

log | tan x|]

π

4

π

12

= − log tan( π

12

)

= log cot( π

12

)

.

20.

∫ 2

1

1

ex + e−xdx =

∫ 2

1

ex

e2x + 1dx =

∫ e2

e

1

u2 + 1du =

[

tan−1 u]e2

e

= tan−1 e2 − tan−1 e = tan−1 e2 − e

1 + e3.

21

∫ b

a

1

xsec2 log x dx =

∫ log b

log a

sec2 u du =[

tan u]log b

log a= tan log b− tan log a.

9

22.∫ 1

0

sin−1 x+ cos−1 x dx =

∫ 1

0

π

2dx =

π

2.

Draw the graph!

23.

∫ π

4

π

3

sec x dx =[

log | sec x+ tanx|]

π

4

π

3

= log(√2 + 1)− log(2 +

√3) = log

(√2 + 1√3 + 2

)

= − log

(√3 + 2√2 + 1

)

,

which is negative, because of the unusual limits on the integral.

24.

∫ 4

1

x3 − x

x3

2

dx =

∫ 4

1

x3

2 − x−1

2 dx =[2

5x

5

2 − 2x1

2

]4

1=

2

5(32 − 1)− 2(2− 1) =

52

5.

25.

∫ π2

4

π2

9

sin√x√

xdx =

∫ π

2

π

3

2 sin u du =[

− 2 cos u]

π

2

π

3

= −2

(

0− 1

2

)

= 1.

26.∫ π

2

0

1

1 + cosxdx =

∫ π

2

0

1

1 + sinxdx = 1.

(Done earlier!)

27.

∫ e

1

x+ 49

x− 49dx =

∫ e

1

1 +98

x− 49dx =

[

x+ 98 log |x− 49|]e

1= e− 1 + 98 log

∣

∣

∣

e− 49

1− 49

∣

∣

∣

= e− 1 + 98 log

(

49− e

48

)

.

Draw the graph!

10

28.

∫ 7

5

2log x dx =

∫ 7

5

xlog 2 dx =[ xlog 2+1

log 2 + 1

]7

5=

1

log 2 + 1

(

7log 2+1 − 5log 2+1)

.

29.

∫ π

2

0

1

sec x+ tanx sinxdx =

∫ π

2

0

cos x

1 + sin2 xdx =

∫ 1

0

1

1 + u2du =

[

tan−1 u]1

0=

π

4.

30.

∫ e

1

1− log x

x2dx =

∫ 1

0

1− t

e2tet dt =

∫ 1

0

(1− t)e−t dt =[

te−t]1

0= e−1.

31.∫ π

3

0

sec4 x tan x dx =

∫ 2

1

u3 du =1

24.

32.∫ 1

0

x2(x3 + 1)4 dx =

∫ 2

1

1

3u4 du =

[u5

15

]2

1=

31

15.

33.

∫ 2

π

0

sin2 x dx =

∫ 2

π

0

1

2− 1

2cos 2x dx =

[1

2x− 1

4sin 2x

]2

π

0=

1

π− 1

4sin

(

4

π

)

.

34.∫ π

e

cos 3x

4 cos3 x− 3 cos xdx =

∫ π

e

1 dx = π − e.

(You have to know that cos 3x = 4cos3 x− 3 cos x. Try to prove it.)

35.

2

π

∫ π

4

π

8

tan−1 θ + cot−1 θ dθ =2

π

∫ π

4

π

8

1 dθ =2

π· π8=

1

4.

Unusual. tan−1 and cot−1 are usually applied to x, not θ.

11

36.∫ π

4

0

etan x

cos2 xdx =

∫ 1

0

eu du = e− 1.

37.

∫

√π

2

√π

6

x cos x2 dx =

∫ π

2

π

6

1

2cos u du =

[1

2sinu

]π

2

π

6

=1

2

(

1− 1

2

)

=1

4.

38.∫ sec2( π

4 )−1

1

ex dx =

∫ 1

1

ex dx = 0.

39.∫ log 3

0

ex

1 + exdx =

∫ 4

2

1

udu = log 4− log 2 = log 2.

40.

∫ 0

−1

2x+ 1

x2 + 2x+ 2dx =

∫ 1

0

2u− 1

u2 + 1du =

[

log(u2 + 1)− tan−1 u]1

0= log 2− π

4.

41.

∫ 12

4

1

(4 + x)√x

dx =

∫

√

12

2

1

(4 + u2)u2u du =

∫

√

12

2

2

4 + u2du =

[

tan−1(u

2

) ]

√

12

2

= tan−1√3− tan−1 1 =

π

3− π

4=

π

12.

42.∫ 3

1

2x2 − x+ 1

x− 2dx

does not exist, because of the discontinuity at x = 2.

43.

∫

√

3

−

√

3

−1√4− x2

dx =[

− sin−1(x

2

) ]

√

3

−

√

3= −

(π

3−(

−π

3

))

= −2π

3.

12

44.∫ π

4

0

sin2 x dx =[1

2x− 1

4sin 2x

]π

4

0=

π

8− 1

4.

45.

∫ 1

0

√x

1 + xdx =

∫ 1

0

u

1 + u22u du =

∫ 1

0

2u2

u2 + 1du =

∫ 1

0

2− 2

u2 + 1du

=[

2u− 2 tan−1 u]1

0= 2− π

2.

46.∫ 2

1

(x+ 1)−2 dx =[

− 1

x+ 1

]2

1= −1

3+

1

2=

1

6.

47.

∫ 4

2

(√x− 1)2 dx =

∫ 2

√

2

(u− 1)22u du =

∫ 2

√

2

2u3 − 4u2 + 2u du =[1

2u4 − 4

3u3 + u2

]2

√

2

=1

2(16− 4)− 4

3(8− 2

√2) + (4− 2) = 6− 4

3(8− 2

√2 + 2 = −8

3+

8

3

√2 =

8

3(√2− 1).

48.

∫

−1

2

x3 − 2x2 + 3x− 4 dx =[1

4x4 − 2

3x3 +

3

2x2 − 4x

]

−1

2

=1

4

(

(−1)4 − 24)

− 2

3

(

(−1)3 − 23)

+3

2

(

(−1)2 − 22)

− 4 (−1− 2)

=1

4×−15− 2

3×−9 +

3

2×−3− 4×−3 = −15

4+ 6− 9

2+ 12 = 18− 15

4− 18

4

= 18− 33

4=

39

4.

13

Quarter–Finals

1.

∫

1√sin3 x cos x

dx =

∫

1√

tan3 x/ sec4 xdx =

∫

sec2 x

(tan x)3

2

dx = −2(tan x)−1

2 + C

= − 2√tanx

+ C.

2.

∫

√tanx

sinx cos xdx =

∫

√tanx

tan x/ sec2 xdx =

∫

secx

(tan x)1

2

dx = 2(tan x)1

2 +C

= 2√tan x+ C.

3.

∫

sin5 x

cos xdx =

∫

sinx(1− cos2 x)2

cos xdx =

∫

sinx(1− 2 cos2 x+ cos4 x)

cos xdx

=

∫

tanx− 2 cos x sinx+ cos3 x sinx dx = log | sec x|+ cos2 x− 1

4cos4 x+ C.

4.

∫

cos√x dx =

∫

2u cos u du = 2u sinu+2cosu+C = 2√x sin

√x+2cos

√x+C.

5.

∫

3x− 1

(x2 + 1)(x2 + 2)dx =

∫

(3x− 1)

(

1

x2 + 1− 1

x2 + 2

)

dx =

∫

3x− 1

x2 + 1− 3x− 1

x2 + 2dx

=3

2log(x2 + 1)− tan−1 x− 3

2log(x2 + 2) +

1√2tan−1

(

x√2

)

+ C.

6.∫

x

x4 + 1dx =

∫

1

2

1

u2 + 1du =

1

2tan−1 u+ C =

1

2tan−1 x2 + C.

14

7.∫

log cos x

cot xdx =

∫

− log u

udu =

∫

−v dv = −1

2v2 + C = −1

2(log |u|)2 + C

= −1

2(log | cos x|)2 + C.

8.

∫

sin5 x cos6 x dx =

∫

cos6 x(1−cos2 x)2 sinx dx = −1

7cos7 x+

2

9cos9 x− 1

11cos11 x+C.

9.∫

1√x2 + 2x

dx =

∫

1√

(x+ 1)2 − 1dx =

∫

1√u2 − 1

du

=

∫

1√sec2 θ − 1

sec θ tan θ dθ =

∫

sec θ dθ = log | sec θ + tan θ|+ C

= log |u+√

u2 − 1|+ C = log |x+ 1 +√

x2 + 2x|+ C.

10.∫

(3 + 2x) log x dx =

∫

(3 + 2et)tet dt =

∫

3tet + 2te2t dt

= 3tet − 3et + te2t − 1

2e2t + C = 3x log |x| − 3x+ x2 log |x| − 1

2x2 + C

= (x2 + 3x) log |x| − 1

2x2 − 3x+ C.

11.

∫

12

(x2 + 4)(x2 + 16)dx =

∫

1

x2 + 4− 1

x2 + 16dx =

1

2tan−1

(x

2

)

−1

4tan−1

(x

4

)

+C.

12.

If for the moment we pretend√x2 = x, we find

∫

√

x2 + x4 dx =

∫

x√

1 + x2 dx =

∫

1

2u

1

2 du =1

3u

3

2 + C =1

3(1 + x2)

3

2 + C

=1

3x3(x2 + x4)

3

2 +C.

15

Luckily, since this function is odd (apart from the C), it is correct.

13.

∫

√

1− x2 dx =

∫

√

1− sin2 θ cos θ dθ =

∫

cos2 θ dθ =

∫

1

2+

1

2cos 2θ dθ

=1

2+

1

4sin 2θ + C =

1

2θ +

1

2sin θ cos θ + C =

1

2sin−1 x+

1

2x√

1− x2 + C.

Actually, a similar comment applies to this integral as to the previous one!

14.

∫

1

(4 + x2)3

2

dx =

∫

1

(4 + 4 tan2 θ)3

2

· 2 sec2 θ dθ =

∫

1

8 sec3 θ· 2 sec2 θ dθ

=

∫

1

4cos θ dθ =

1

4sin θ + C =

1

4

x√4 + x2

+ C.

15.

∫

1√e2x − 1

dx =

∫

ex

ex√e2x − 1

dx =

∫

1

u√u2 − 1

du

=

∫

1

sec θ tan θsec θ tan θ dθ =

∫

1 dθ = θ + C = sec−1 u+ C

= sec−1 ex + C.

Alternatively, = tan−1√e2x − 1 + C.

16.

∫

cos4 x dx =

∫(

1

2+

1

2cos 2x

)2

dx =

∫

1

4

(

1 + 2 cos 2x+ cos2 2x)

dx

=

∫

1

4

(

1 + 2 cos 2x+1

2+

1

2cos 4x

)

dx =

∫

3

8+

1

2cos 2x+

1

8cos 4x dx

=3

8x+

1

4sin 2x+

1

32sin 4x+ C =

3

8x+

1

2sinx cos x+

1

16sin 2x cos 2x+ C

=3

8x+

1

2sinx cos x+

1

8sinx cos x(2 cos2 x− 1) + C

=3

8x+

3

8sinx cos x+

1

4sinx cos3 x+ C.

16

17.

∫

1√9 + 16x− 4x2

dx =

∫

1

2√

94+ 4x− x2

dx =

∫

1

2√

254− (x− 2)2

dx

=

∫

1

2

√

(

52

)2 − (x− 2)2dx =

1

2sin−1

(

2

5(x− 2)

)

+ C.

18.

∫

sin log x dx =

∫

eu sinu du = eu(sinu−cos u)+C = x(sin log x−cos log x)+C.

(See question 9 of the Qualifying Round.)

19.

∫

2x tan−1 x dx = x2 tan−1 x−∫

x2

1 + x2dx = x2 tan−1 x−

∫

1− 1

1 + x2dx

= x2 tan−1 x− x+ tan−1 x+ C = (1 + x2) tan−1 x− x+ C.

20.

∫

x3ex2

(x2 + 1)2dx =

∫

1

2

ueu

(u+ 1)2du =

1

2

∫

(v − 1)ev−1

v2dv =

1

2e

∫

(v − 1)ev

v2dv

=1

2e

{

(v − 1)ev · −1

v−∫

−1

v· vev dv

}

=1

2e

{

−ev +ev

v+

∫

ev dv

}

=1

2e

ev

v+ C

=1

2e

eu+1

u+ 1+ C =

1

2

eu

u+ 1+ C =

1

2

ex2

x2 + 1+ C =

ex2

2(x2 + 1)+C.

21.

∫

(x− 1)(x+ 1)11 dx =

∫

u11(u− 2) du =

∫

u12 − 2u11 du =1

13u13 − 1

6u12 + C

=1

13(x+ 1)13 − 1

6(x+ 1)12 + C =

1

78(x+ 1)12 (6(x+ 1)− 13) + C

=1

78(x+ 1)12(6x− 7) + C..

17

22.

∫

1

1 + x1

4

dx =

∫

1

1 + u4u3 du =

∫

4

v(v − 1)3 dv =

∫

4(v2 − 3v + 3− 1

v) dv

= 4(1

3v3 − 3

2v2 + 3v − log |v|) + C =

4

3(u+ 1)3 − 6(u+ 1)2 + 12(u+ 1)− 4 log |u+ 1|+ C

=4

3u3 − 2u2 + 4u− 4 log |u+ 1|+ C =

4

3x

3

4 − 2x1

2 + 4x1

4 − 4 log |1 + x1

4 |+ C.

23.

∫

sin5(x−1) cos(x−1)

x2dx =

∫

− sin5 u cos u du = −1

6sin6 u+C = −1

6sin6(x−1)+C

24.

∫

2(√

2 cos(

x+ π4

))2dx =

∫

sec2(

x+π

4

)

dx = tan(

x+π

4

)

+ C.

25.

∫

(

sin−1 x)2

dx =

∫

u2 cos u du = u2 sinu+ 2u cos u− 2 sin u+ C

= x(

sin−1 x)2

+ 2√

1− x2 sin−1 x− 2x+ C.

18

Semi–Finals

1.

∫

sec2 x

sec x+ tan xdx =

∫

sec2 x (sec x− tan x) dx =

∫

sec3 x dx− 1

2tan2 x+ C.

The notoriously ugly

∫

sec3 x dx can be calculated by integration by parts.

∫

sec3 θ dθ =1

2sec θ tan θ +

1

2log | sec θ + tan θ|+ C.

So our integral

=1

2sec θ tan θ +

1

2log | sec θ + tan θ| − 1

2tan2 θ + C

=1

2tan θ(secθ − tan θ) +

1

2log | sec θ + tan θ|+C.

2.

∫

exx

(log x+1)x2x dx =

∫

ueu du = ueu−eu+C = xxexx−ex

x

+C = exx

(xx−1)+C.

3.

∫

2x− 9√x+ 9

(x− 3√x)

1

3

dx =

∫

2u2 − 9u+ 9

(u2 − 3u)1

3

2u du =

∫

4u3 − 18u2 + 18u

(u2 − 3u)1

3

du

=

∫

(u2 − 3u)(4u− 6)

(u2 − 3u)1

3

du =

∫

(u2 − 3u)2

3 2(2u − 3) du =

∫

2v2

3 dv

=6

5v

5

3 + C =6

5(u2 − 3u)

5

3 + C =6

5(x− 3

√x)

5

3 + C.

4.

∫

x5

√1 + x2

dx =

∫

1

2

(u− 1)2√u

du =

∫

1

2

(

u3

2 − 2u1

2 + u−1

2

)

du

=1

5u

5

2 − 2

3u

3

2 + u1

2 +C =1

15u

1

2 (3u2 − 10u+ 15) + C

=1

15(3(x2 + 1)2 − 10(x2 + 1) + 15)

√

1 + x2 + C =1

15(3x4 − 4x2 + 8)

√

1 + x2 + C.

19

5.

∫

log(1 + log x)

xdx =

∫

log u du = u log u− u+ C

= (1 + log x) log(1 + log x)− log x+C.

6.

∫

x1

2

x1

3 + x1

4

dx =

∫

u6

u4 + u312u11 du =

∫

12u14

u+ 1du =

∫

12(u14 − 1) + 12

u+ 1du

=

∫

12(

u13 − u12 + u11 − u10 + u9 − u8 + u7 − u6 + u5 − u4

+ u3 − u2 + u− 1 +1

u+ 1

)

du

=6

7u14 − 12

13u13 + u12 − 12

11u11 +

6

5u10 − 4

3u9 +

3

2u8 − 12

7u7 + 2u6 − 12

5u5

+ 3u4 − 4u3 + 6u2 − 12u+ 12 log |u+ 1|+ C

Now set u = x1

12 .

7.

∫

x3 + 3x2 + 3x

x4 + 4x3 + 6x2 + 4x+ 1dx =

∫

(x+ 1)3 − 1

(x+ 1)4dx =

∫

1

x+ 1− 1

(x+ 1)4dx

= log |x+ 1|+ 1

3(x+ 1)3+C.

8.

∫

secx cosecx dx =

∫

1

cos x sinxdx =

∫

2

sin 2xdx =

∫

2cosec2x dx

= − log |cosec 2x+ cot 2x|+ C.

9.

∫

1

x+ x4dx =

∫

1

x(x+ 1)(x2 − x+ 1)dx =

∫

A

x+

B

x+ 1+

Cx+D

x2 − x+ 1dx

= A log |x|+B log |x+ 1|+ C

2log(x2 − x+ 1) +

(

D +C

2

)

sin−1

(

2√3

(

x− 1

2

))

+K,

20

where A, B, C and D are given by

A(x+ 1)(x2 − x+ 1) +Bx(x2 − x+ 1) + (Cx+D)x(x+ 1) = 1

(for all x).

We have

A+B + C = 0, −B +C +D = 0, B +D = 0, A = 1,

so

A = 1, B = −1

3, C = −2

3, D =

1

3,

and the integral is

= log |x| − 1

3log |x+ 1| − 1

3log(x2 − x+ 1) +K = log |x| − 1

3log |x3 + 1|+K

=1

3log∣

∣

∣

x3

x3 + 1

∣

∣

∣+K =

1

3log∣

∣

∣

x4

x+ x4

∣

∣

∣+K.

10.∫

cosec x dx = − log |cosec x+ cot x|+ C.

(Pure bookwork.)

11.∫

cos sin sinx cos sinx cos x dx = sin sin sinx+ C.

(Obvious!)

12.∫

x√1− 2x− x2

dx =

∫

x√

2− (x+ 1)2dx =

∫

u− 1√2− u2

du

= −√

2− u2 − sin−1

(

u√2

)

+ C = −√

1− 2x− x2 − sin−1

(

x+ 1√2

)

+ C.

13.∫

x3 log(5x) dx =

∫

1

54u3 log u du =

∫

1

54ve4v dv =

1

54

(

1

4ve4v − 1

16e4v)

+ C

=1

54

(

1

4u4 log u− 1

16u4

)

+ C =1

4x4 log 5x− 1

16x4 + C.

21

14.

∫(

log x

x

)2

dx =

∫

( u

eu

)2

eu du =

∫

u2e−u du

= −u2e−u − 2ue−u − 2e−u + C = − (log x)2

x− 2

log x

x− 2

x+ C.

15. If n is an integer, n ≥ 2,

∫

1

xn + xdx =

∫

1

x(xn−1 + 1)dx =

∫

xn−2

xn−1(xn−1 + 1)dx

=

∫

xn−2

(

1

xn−1− 1

xn−1 + 1

)

dx

=

∫

1

x− xn−2

xn−1 + 1dx = log |x| − 1

n− 1log |xn−1 + 1|+C.

22

Finals

1.∫

sin2 x cos4 x dx =

∫(

1

2sin 2x

)2(1

2+

1

2cos 2x

)

dx =1

8

∫

sin2 2x+ sin2 2x cos 2x dx

=1

8

∫

1

2− 1

2cos 4x+ sin2 2x cos 2x dx =

1

16x− 1

64sin 4x+

1

48sin3 2x+ C.

This solution was indicated to me by one of the competitors.

2.∫

4x2 − 15x+ 29

(x− 5)(x2 − 4x+ 13)dx =

∫

A

x− 5+

Bx+ C

(x− 2)2 + 32dx

= A log |x− 5|+ B

2log(x2 − 4x+ 13) +

2B + C

3tan−1

(

x− 2

3

)

+K,

where A, B and C are given by

A(x2 − 4x+ 13) + (Bx+ C)(x− 5) = 4x2 − 15x+ 29

(for all x).

We have

A+B = 4, −4A− 5B +C = −15, 13A− 5C = 29,

so

A = 3, B = 1, C = 2

and the integral is

= 3 log |x− 5|+ 1

2log(x2 − 4x+ 13) +

4

3tan−1

(

x− 2

3

)

+K.

3.∫ √

1 + sinx cot x dx =

∫

√1 + sinx

sinxcos x dx =

∫

√1 + u

udu

=

∫

v

v2 − 12v dv =

∫

2v2

v2 − 1dv =

∫

2 +2

v2 − 1dv =

∫

2 +

(

1

v − 1− 1

v + 1

)

dv

= 2v + log∣

∣

∣

v − 1

v + 1

∣

∣

∣+C = 2

√1 + u+ log

∣

∣

∣

√1 + u− 1√1 + u+ 1

∣

∣

∣+ C

= 2√1 + sinx+ log

∣

∣

∣

√1 + sinx− 1√1 + sinx+ 1

∣

∣

∣+ C.

23

4.∫

sin 2x

1 + 2 sin2 xdx =

∫

sin 2x

2− cos 2xdx =

∫

−1

2

1

2− udu =

1

2log |2− u|+ C

=1

2log |2− cos 2x|+ C =

1

2log(2− cos 2x) + C.

5.∫

x4

1− x2dx =

∫

− x4

x2 − 1dx =

∫

−x4 − 1 + 1

x2 − 1dx =

∫

−(

x2 + 1 +1

x2 − 1

)

dx

=

∫

−x2 − 1 +1

1− x2dx =

∫

−x2 − 1 +1

2

(

1

1− x+

1

1 + x

)

dx

= −1

3x3 − x+

1

2log∣

∣

∣

1 + x

1− x

∣

∣

∣+ C

6.

∫√1− x

1−√x

dx =

∫

√

1− sin2 θ

1− sin θ2 sin θ cos θ dθ =

∫

cos2 θ

1− sin θ2 sin θ dθ

=

∫

(1 + sin θ)2 sin θ dθ =

∫

2 sin θ + 1− cos 2θ dθ = −2 cos θ + θ − 1

2sin 2θ +C

= θ − 2 cos θ − sin θ cos θ +C = sin−1√x− 2

√1− x−

√x√1− x+ C.

7.∫

ecot x+2 log cosec x dx =

∫

ecot xcosec2x dx =

∫

−eudu = −eu + C = −ecot x + C.

8.

∫

xesin−1 x

√1− x2

dx =

∫

eθ sin θ dθ =1

2(sin θ − cos θ)eθ + C

=1

2

(

x−√

1− x2

)

esin−1 x + C

9.∫

3 sin x+ 2cos x

2 sin x+ 3cos xdx =

∫ 1213(2 sin x+ 3cos x)− 5

13(2 cos x− 3 sin x)

2 sin x+ 3cos xdx

=

∫

12

13− 5

13

2 cos x− 3 sinx

2 sin x+ 3cos xdx =

12

13x− 5

13log |2 sin x+ 3cos x|+ C.

24

10.

∫

1

sinx+ sin 2xdx =

∫

1

sinx+ 2 sinx cos xdx =

∫

1

sinx(1 + 2 cos x)dx

=

∫

1

(1− cos2 x)(1 + 2 cos x)sinx dx =

∫

− 1

(1− u2)(1 + 2u)du

=

∫

1

2

1

1 + u− 1

6

1

1− u− 4

3

1

1 + 2udu

=1

2log |1 + u|+ 1

6log |1− u| − 2

3log |1 + 2u|+C

=1

2log |1 + cos x|+ 1

6log |1− cos x| − 2

3log |1 + 2 cos x|+ C

=1

6log

(

(1 + cos x)3(1− cos x)

(1 + 2 cos x)4

)

+ C =1

6log

(

sin2 x(1 + cos x)2

(1 + 2 cos x)4

)

+ C

=1

3log∣

∣

∣

sinx(1 + cos x)

(1 + 2 cos x)2

∣

∣

∣+ C

25

Tie–breaker questions

1.

Let I =

∫ π

2

0

√sinx√

sinx+√cos x

dx.

Then

I =

∫ π

2

0

√

sin(

π2− x)

√

sin(

π2− x)

+√

cos(

π2− x)

dx =

∫ π

2

0

√cos x

√cos x+

√sinx

dx,

2I =

∫ π

2

0

1 dx =π

2, so I =

π

4.

2.∫ e

1

2

1

sin−1 log x

xdx =

∫ 1

2

0

sin−1 u du =[

u sin−1 u]

1

2

0−∫ 1

2

0

u√1− u2

du

=π

12+[

√

1− u2

]1

2

0=

π

12+

√3

2− 1.

3.∫ 1

0

x7 − 1

log xdx =

∫ 1

0

∫ 7

0

xy dy dx =

∫ 7

0

∫ 1

0

xy dx dy =

∫ 7

0

[ xy+1

y + 1

]1

0dy

=

∫ 7

0

1

y + 1dy =

[

log(y + 1)]7

0= log 8.

Clearly not suitable for first–year students.

4.

Let I =

∫ π

2

0

1

1 + tanπ xdx.

Then I =

∫ π

2

0

1

1 + tanπ(

π2− x) dx =

∫ π

2

0

1

1 + cotπ xdx =

∫ π

2

0

tanπ x

tanπ x+ 1dx,

and 2I =

∫ π

2

0

1 dx =π

2, I =

π

4.

5.∫ π

6

0

sec3 2θ dθ =

∫ π

3

0

1

2sec3 u du =

[1

4sec θ tan θ +

1

4log | sec θ + tan θ|

]π

3

0

=

√3

2+

1

4log(2 +

√3).