009B Sample Midterm 1, Problem 1
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Evaluate the indefinite and definite integrals.
a)
∫
x
2
1
+
x
3
d
x
{\displaystyle \int x^{2}{\sqrt {1+x^{3}}}dx}
b)
∫
π
4
π
2
cos
(
x
)
sin
2
(
x
)
d
x
{\displaystyle \int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\cos(x)}{\sin ^{2}(x)}}dx}
Foundations:
Solution:
(a)
Step 1:
We need to use substitution. Let
u
=
1
+
x
3
{\displaystyle u=1+x^{3}}
. Then,
d
u
=
3
x
2
d
x
{\displaystyle du=3x^{2}dx}
and
d
u
3
=
x
2
d
x
{\displaystyle {\frac {du}{3}}=x^{2}dx}
.
Therefore, the integral becomes
1
3
∫
u
d
u
{\displaystyle {\frac {1}{3}}\int {\sqrt {u}}du}
.
Step 2:
We now have:
∫
x
2
1
+
x
3
d
x
=
1
3
∫
u
d
u
=
2
9
u
3
2
+
C
=
2
9
(
1
+
x
3
)
3
2
+
C
{\displaystyle \int x^{2}{\sqrt {1+x^{3}}}dx={\frac {1}{3}}\int {\sqrt {u}}du={\frac {2}{9}}u^{\frac {3}{2}}+C={\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C}
.
(b)
Step 1:
Step 2:
Final Answer:
(a)
2
9
(
1
+
x
3
)
3
2
+
C
{\displaystyle {\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C}
(b)
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