Quiz 1 Friday, August 30
2
1 Find the antiderivative of x23 − 2 sin( x2 ) + sec 3(2x)
Solution.
First, we can simplify the expression to get
2
2
− 2 sin( x2 ) + sec 3(2x) = 2x−3 − 2 sin( x2 ) + 13 sec2 (2x).
x3
Method 1:
−3+1
Note that the antiderivative of x−3 is x−3+1 + c =
the antiderivative of sin( x2 ) is −
cos( x2 )
1
2
+c =
x−2
+ c,
−2
x
−2 cos( 2 ) + c and
the anti-
derivative of sec2 (2x) is tan(2x)
+ c.
2
2
2
So the antiderivative of x3 − 2 sin( x2 ) + sec 3(2x) is
−2
+c
2 · x−2 − 2 · (−2 cos( x2 )) + 13 · tan(2x)
2
which can be simplified as
x
1
−x−2 + 4 cos( )) + tan(2x) + c.
2
6
One can verify the answer by taking the derivative of −x−2 +4 cos( x2 ))+
1
tan(2x) + c to get
6
x
1
d
− x−2 + 4 cos( )) + tan(2x) + c
dx
2
6
1
x 1
= − · (−2)x−3 + 4 · (− cos( ) · ) + sec2 (2x) · 2 + 0
2 2
6
x
1
= 2x−3 − 2 cos( ) + sec2 (2x).
2
3
Method 2: We can use indefinite integral to find the antiderivative
of a function.
2
Since x23 − 2 sin( x2 ) + sec 3(2x) = 2x−3 − 2 sin( x2 ) + 13 sec2 (2x), we have
Z x
sec2 (2x) 2
−
2
sin(
)
+
dx
x3
2
3
Z x
1
2x−3 − 2 sin( ) + sec2 (2x))dx
=
2
3
Z
Z
Z
x
1
−3
= 2 x dx − 2 sin( )dx +
sec2 (2x)dx
2
3
x
−2
)
cos(
x
1 tan(2x)
=2·
− 2 · (− 1 2 ) + ·
+c
−2
3
2
2
x
1
= −x−2 + 4 cos( )) + tan(2x) + c
2
6
So the antiderivative of x23 − 2 sin( x2 ) +
−x−2 + 4 cos( x2 )) + 16 tan(2x) + c.
sec2 (2x)
3
MATH 1760 : page 1 of 1
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