17B Discussion Sheet 0
1. Find general antiderivative for the given functions. Assume that all functions are defined for all positive real number.
(a) f (x) = 4x5
Ans:
Z
Z
f (x) dx =
(b) f (x) = 5x4
Ans:
Z
Z
f (x) dx =
(c) f (x) = x−1
Ans:
4x dx = 4
Z
4
5x dx = 5
Z
Z
f (x) dx =
(d) f (x) = sin(3x)
Ans:
Z
5
Z
x
Z
f (x) dx =
−1
4
x5 dx = x5 + C.
5
5
x4 dx = x5 + C = x5 + C.
5
Z
dx =
1
dx = ln x + C.
x
1
sin(3x) dx = − cos(3x) + C.
3
(e) f (x) = cos(5x + 1)
Ans:
Z
Z
1
f (x) dx = cos(5x + 1) dx = sin(5x + 1) + C.
5
(f) f (x) = 5 sec2 (2x − 1)
Ans:
Z
Z
5
f (x) dx = 5 sec2 (2x − 1) dx = tan(2x − 1) + C.
2
√
3
(g) f (x) = 2 x+x
x2
Ans:
Z
Z √
Z
Z
2 x + x3
x2
−3/2
−1/2
f (x) dx =
+C
dx
=
2
x
dx
+
x
dx
=
−4x
+
x2
2
(h) f (x) = e3x
Ans:
Z
Z
f (x) dx =
2
1
e3x dx = e3x + C.
3
(i) f (x) = 2xex
Ans: Let u = x2 , then du = 2x dx. Therefore,
Z
Z
Z
2
x2
f (x) dx = 2xe dx = eu du = eu + C = ex + C.
1
Solutions to problems 2 and 3 from discussion sheet #0
January 6, 2015
2.
dy
dx
d
= etan(3x) dx
tan(3x) = etan 3x sec2 (3x)3
R
• b) f (x) = etan(5x) sec2 (5x) ⇒ F (x) = etan(5x) sec2 (5x)dx. Using ”u-substitution”
R u
u
tan(5x)
with u = tan(5x) we obtain: F (x) = e5 du = e5 + C = e 5 + C.
• a) y = etan(3x) ⇒
3.
dy
• a) y = sin2 (3x) ⇒ dx
= 2 sin(3x) cos(3x)3 = 6 sin(3x) cos(3x).
dy
5
y = sin (3x) ⇒ dx = 5 sin4 (3x) cos(3x)3 = 15 sin4 (3x) cos(3x).
R
• b) f (x) = 15 cos(5x) sin(5x) ⇒ F (x) = 15 cos(5x) sin(5x)dx = 32 sin2 (5x) + C.
R
• c) f (x) = 15 cos(5x) sin2 (5x) ⇒ F (x) = 15 cos(5x) sin2 (5x)dx = sin3 (5x) + C.
1
17B Discussion Sheet 0
Problem 4
(a) Take derivative by Chain rule to get:
dy
3x2 + 3
= 3
dx
x + 3x + 6
(b) We can get a clue from problem (a), where the numerator of the rational function is
related to the derivative of its denominator. In particular, (x4 +8x2 +16)0 = 4(x3 +4x),
so we have the antiderivative:
F (x) =
(c) Similarly with (b), since tan x =
1
ln(x4 + 8x2 + 16)
4
sin x
,
cos x
where (cos x)0 = − sin x, we have:
F (x) = − ln(cos x)
(d) we can use the table to get:
1
F (x) = − ln(cos 7x)
7
(e) Similarly with (d):
1
F (x) = − ln(cos(7x + 9))
7
1
5. Find a function f (x) whose antiderivative is given as y.
Solution: It suffices to find the derivatives for the given functions.
(a) y = xsin x
Note that ln y = sin x ln x. So, taking derivative with respect to x at each side
gives
sin x
1 0
y = cos x ln x +
.
y
x
So,
sin x
y = cos x ln x +
xsin x .
x
0
x2
(b) y = (x3 ) ee
Applying product rule and chain rule gives
x2
y 0 = 3x2 ee
2
x2
+ 2x4 ex ee .
(c) y = 12 ex cos x
Applying product rule gives
1
y 0 = ex (cos x − sin x) .
2
(d) y = ln (cos x)
Applying chain rule gives
y 0 = − tan x.
(e) y = sin(g(x))
Applying chain rule gives
y 0 = cos(g(x))g 0 (x).
(f) y = ln(ln x)
Applying chain rule gives
y0 =
1
1
.
x ln x
(g) y = ex
2 +x
Applying chain rule gives
y 0 = (2x + 1)ex
(g) y = ln(x + 1) +
2 +x
1
x+1
Applying chain rule gives
y0 =
2
x
.
(x + 1)2
.