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Evaluate the following:
Find the general antiderivative of the following:
Find the anti-derivative or compute the integral depending on whether the integral is indefinite or definite.
13.
∫
0
π
/
2
sin
(
x
)
cos
(
x
)
d
x
{\displaystyle \int _{0}^{\pi /2}\sin(x)\cos(x)\,dx}
Notice that
sin
(
2
x
)
=
2
sin
(
x
)
cos
(
x
)
1
2
sin
(
2
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle {\begin{aligned}\sin(2x)&=2\sin(x)\cos(x)\\{\frac {1}{2}}\sin(2x)&=\sin(x)\cos(x)\end{aligned}}}
By this,
∫
0
π
/
2
sin
(
x
)
cos
(
x
)
d
x
=
∫
0
π
/
2
1
2
sin
(
2
x
)
d
x
{\displaystyle \int _{0}^{\pi /2}\sin(x)\cos(x)\,dx=\int _{0}^{\pi /2}{\frac {1}{2}}\sin(2x)\,dx}
.
Let
u
=
2
x
,
d
u
=
2
d
x
⟹
1
2
d
u
=
d
x
{\displaystyle u=2x,\qquad du=2\,dx\implies {\frac {1}{2}}du=dx}
.
u
(
0
)
=
0
,
u
(
π
/
2
)
=
π
{\displaystyle u(0)=0,\qquad u(\pi /2)=\pi }
Then,
∫
0
π
/
2
1
2
sin
(
2
x
)
d
x
=
1
2
∫
0
π
1
2
sin
(
u
)
d
u
=
−
1
4
(
cos
(
π
)
−
cos
(
0
)
)
=
1
2
{\displaystyle {\begin{aligned}\int _{0}^{\pi /2}{\frac {1}{2}}\sin(2x)\,dx&={\frac {1}{2}}\int _{0}^{\pi }{\frac {1}{2}}\sin(u)\,du\\&=-{\frac {1}{4}}\left(\cos(\pi )-\cos(0)\right)\\&=\mathbf {\frac {1}{2}} \end{aligned}}}
Notice that
sin
(
2
x
)
=
2
sin
(
x
)
cos
(
x
)
1
2
sin
(
2
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle {\begin{aligned}\sin(2x)&=2\sin(x)\cos(x)\\{\frac {1}{2}}\sin(2x)&=\sin(x)\cos(x)\end{aligned}}}
By this,
∫
0
π
/
2
sin
(
x
)
cos
(
x
)
d
x
=
∫
0
π
/
2
1
2
sin
(
2
x
)
d
x
{\displaystyle \int _{0}^{\pi /2}\sin(x)\cos(x)\,dx=\int _{0}^{\pi /2}{\frac {1}{2}}\sin(2x)\,dx}
.
Let
u
=
2
x
,
d
u
=
2
d
x
⟹
1
2
d
u
=
d
x
{\displaystyle u=2x,\qquad du=2\,dx\implies {\frac {1}{2}}du=dx}
.
u
(
0
)
=
0
,
u
(
π
/
2
)
=
π
{\displaystyle u(0)=0,\qquad u(\pi /2)=\pi }
Then,
∫
0
π
/
2
1
2
sin
(
2
x
)
d
x
=
1
2
∫
0
π
1
2
sin
(
u
)
d
u
=
−
1
4
(
cos
(
π
)
−
cos
(
0
)
)
=
1
2
{\displaystyle {\begin{aligned}\int _{0}^{\pi /2}{\frac {1}{2}}\sin(2x)\,dx&={\frac {1}{2}}\int _{0}^{\pi }{\frac {1}{2}}\sin(u)\,du\\&=-{\frac {1}{4}}\left(\cos(\pi )-\cos(0)\right)\\&=\mathbf {\frac {1}{2}} \end{aligned}}}
14.
∫
0
π
/
4
tan
(
x
)
d
x
{\displaystyle \int _{0}^{\pi /4}\tan(x)\,dx}
.
Rewrite the integral into an equivalent form to help us find the substitution:
∫
0
π
/
4
tan
(
x
)
d
x
=
∫
0
π
/
4
sin
(
x
)
cos
(
x
)
d
x
{\displaystyle \int _{0}^{\pi /4}\tan(x)\,dx=\int _{0}^{\pi /4}{\frac {\sin(x)}{\cos(x)}}\,dx}
Let
u
=
cos
(
x
)
,
−
d
u
=
sin
(
x
)
d
x
{\displaystyle u=\cos(x),\qquad -du=\sin(x)\,dx}
.
u
(
0
)
=
1
,
u
(
π
/
4
)
=
1
2
{\displaystyle u(0)=1,\qquad u(\pi /4)={\frac {1}{\sqrt {2}}}}
.
Apply all this information to find the original integral:
∫
0
π
/
4
sin
(
x
)
cos
(
x
)
d
x
=
−
∫
1
1
/
2
1
u
d
u
=
∫
1
/
2
1
1
u
d
u
=
ln
(
1
)
−
ln
(
1
2
)
=
0
+
ln
(
2
)
=
ln
(
2
)
2
{\displaystyle {\begin{aligned}\int _{0}^{\pi /4}{\frac {\sin(x)}{\cos(x)}}\,dx&=-\int _{1}^{1/{\sqrt {2}}}{\frac {1}{u}}\,du\\&=\int _{1/{\sqrt {2}}}^{1}{\frac {1}{u}}\,du\\&=\ln(1)-\ln \left({\frac {1}{\sqrt {2}}}\right)\\&=0+\ln \left({\sqrt {2}}\right)\\&=\mathbf {\frac {\ln(2)}{2}} \end{aligned}}}
Rewrite the integral into an equivalent form to help us find the substitution:
∫
0
π
/
4
tan
(
x
)
d
x
=
∫
0
π
/
4
sin
(
x
)
cos
(
x
)
d
x
{\displaystyle \int _{0}^{\pi /4}\tan(x)\,dx=\int _{0}^{\pi /4}{\frac {\sin(x)}{\cos(x)}}\,dx}
Let
u
=
cos
(
x
)
,
−
d
u
=
sin
(
x
)
d
x
{\displaystyle u=\cos(x),\qquad -du=\sin(x)\,dx}
.
u
(
0
)
=
1
,
u
(
π
/
4
)
=
1
2
{\displaystyle u(0)=1,\qquad u(\pi /4)={\frac {1}{\sqrt {2}}}}
.
Apply all this information to find the original integral:
∫
0
π
/
4
sin
(
x
)
cos
(
x
)
d
x
=
−
∫
1
1
/
2
1
u
d
u
=
∫
1
/
2
1
1
u
d
u
=
ln
(
1
)
−
ln
(
1
2
)
=
0
+
ln
(
2
)
=
ln
(
2
)
2
{\displaystyle {\begin{aligned}\int _{0}^{\pi /4}{\frac {\sin(x)}{\cos(x)}}\,dx&=-\int _{1}^{1/{\sqrt {2}}}{\frac {1}{u}}\,du\\&=\int _{1/{\sqrt {2}}}^{1}{\frac {1}{u}}\,du\\&=\ln(1)-\ln \left({\frac {1}{\sqrt {2}}}\right)\\&=0+\ln \left({\sqrt {2}}\right)\\&=\mathbf {\frac {\ln(2)}{2}} \end{aligned}}}
15.
∫
1
/
2
1
e
2
x
−
1
2
x
−
1
d
x
{\displaystyle \int _{1/2}^{1}{\frac {e^{\sqrt {2x-1}}}{\sqrt {2x-1}}}\,dx}
.
Let
u
=
2
x
−
1
,
d
u
=
1
2
x
−
1
d
x
{\displaystyle u={\sqrt {2x-1}},\qquad du={\frac {1}{\sqrt {2x-1}}}\,dx}
.
u
(
1
/
2
)
=
0
,
u
(
1
)
=
1
{\displaystyle u(1/2)=0,\qquad u(1)=1}
.
Then,
∫
1
/
2
1
e
2
x
−
1
2
x
−
1
d
x
=
∫
0
1
e
u
d
u
=
e
1
−
e
0
=
e
−
1
{\displaystyle {\begin{aligned}\int _{1/2}^{1}{\frac {e^{\sqrt {2x-1}}}{\sqrt {2x-1}}}\,dx&=\int _{0}^{1}e^{u}\,du\\&=e^{1}-e^{0}\\&=\mathbf {e-1} \end{aligned}}}
Let
u
=
2
x
−
1
,
d
u
=
1
2
x
−
1
d
x
{\displaystyle u={\sqrt {2x-1}},\qquad du={\frac {1}{\sqrt {2x-1}}}\,dx}
.
u
(
1
/
2
)
=
0
,
u
(
1
)
=
1
{\displaystyle u(1/2)=0,\qquad u(1)=1}
.
Then,
∫
1
/
2
1
e
2
x
−
1
2
x
−
1
d
x
=
∫
0
1
e
u
d
u
=
e
1
−
e
0
=
e
−
1
{\displaystyle {\begin{aligned}\int _{1/2}^{1}{\frac {e^{\sqrt {2x-1}}}{\sqrt {2x-1}}}\,dx&=\int _{0}^{1}e^{u}\,du\\&=e^{1}-e^{0}\\&=\mathbf {e-1} \end{aligned}}}
16.
∫
−
3
−
6
8
x
x
2
−
5
d
x
{\displaystyle \int _{-3}^{-{\sqrt {6}}}{\frac {8x}{\sqrt {x^{2}-5}}}\,dx}
.
Let
u
=
x
2
−
5
,
d
u
=
2
x
d
x
{\displaystyle u=x^{2}-5,\qquad du=2x\,dx}
.
u
(
−
3
)
=
4
,
u
(
−
6
)
=
1
{\displaystyle u\left(-3\right)=4,\qquad u(-{\sqrt {6}})=1}
.
Then,
∫
−
3
−
6
8
x
x
2
−
5
d
x
=
4
∫
−
3
−
6
2
x
x
2
−
5
d
x
=
4
∫
4
1
1
u
d
u
=
−
4
∫
1
4
u
−
1
/
2
d
u
=
−
8
(
4
−
1
)
d
u
=
−
8
(
2
−
1
)
=
−
8
{\displaystyle {\begin{aligned}\int _{-3}^{-{\sqrt {6}}}{\frac {8x}{\sqrt {x^{2}-5}}}\,dx&=4\int _{-3}^{-{\sqrt {6}}}{\frac {2x}{\sqrt {x^{2}-5}}}\,dx\\&=4\int _{4}^{1}{\frac {1}{\sqrt {u}}}\,du\\&=-4\int _{1}^{4}u^{-1/2}\,du\\&=-8\left({\sqrt {4}}-{\sqrt {1}}\right)\,du\\&=-8\left(2-1\right)\\&=\mathbf {-8} \end{aligned}}}
Let
u
=
x
2
−
5
,
d
u
=
2
x
d
x
{\displaystyle u=x^{2}-5,\qquad du=2x\,dx}
.
u
(
−
3
)
=
4
,
u
(
−
6
)
=
1
{\displaystyle u\left(-3\right)=4,\qquad u(-{\sqrt {6}})=1}
.
Then,
∫
−
3
−
6
8
x
x
2
−
5
d
x
=
4
∫
−
3
−
6
2
x
x
2
−
5
d
x
=
4
∫
4
1
1
u
d
u
=
−
4
∫
1
4
u
−
1
/
2
d
u
=
−
8
(
4
−
1
)
d
u
=
−
8
(
2
−
1
)
=
−
8
{\displaystyle {\begin{aligned}\int _{-3}^{-{\sqrt {6}}}{\frac {8x}{\sqrt {x^{2}-5}}}\,dx&=4\int _{-3}^{-{\sqrt {6}}}{\frac {2x}{\sqrt {x^{2}-5}}}\,dx\\&=4\int _{4}^{1}{\frac {1}{\sqrt {u}}}\,du\\&=-4\int _{1}^{4}u^{-1/2}\,du\\&=-8\left({\sqrt {4}}-{\sqrt {1}}\right)\,du\\&=-8\left(2-1\right)\\&=\mathbf {-8} \end{aligned}}}
17.
∫
−
3
2
2
e
3
x
−
2
d
x
{\displaystyle \int -{\frac {3}{2}}{\sqrt {\frac {2}{e^{3x-2}}}}\,dx}
.
It may be easier to see what to substitute once the integrand is written in an equivalent form.
∫
−
3
2
2
e
3
x
−
2
d
x
=
∫
−
3
2
2
e
−
1
2
(
3
x
−
2
)
d
x
{\displaystyle \int -{\frac {3}{2}}{\sqrt {\frac {2}{e^{3x-2}}}}\,dx=\int -{\frac {3{\sqrt {2}}}{2}}e^{-{\frac {1}{2}}(3x-2)}\,dx}
From there, it becomes obvious to let
u
=
−
1
2
(
3
x
−
2
)
,
d
u
=
−
3
2
d
x
{\displaystyle u=-{\frac {1}{2}}(3x-2),\qquad du=-{\frac {3}{2}}\,dx}
.
Then,
∫
−
3
2
2
e
−
1
2
(
3
x
−
2
)
d
x
=
∫
2
e
u
d
u
=
2
e
u
+
C
=
2
e
−
1
2
(
3
x
−
2
)
+
C
=
2
e
3
x
−
2
+
C
{\displaystyle {\begin{aligned}\int -{\frac {3{\sqrt {2}}}{2}}e^{-{\frac {1}{2}}(3x-2)}\,dx&=\int {\sqrt {2}}e^{u}\,du\\&={\sqrt {2}}e^{u}+C\\&={\sqrt {2}}e^{-{\frac {1}{2}}(3x-2)}+C\\&=\mathbf {{\sqrt {\frac {2}{e^{3x-2}}}}+C} \end{aligned}}}
It may be easier to see what to substitute once the integrand is written in an equivalent form.
∫
−
3
2
2
e
3
x
−
2
d
x
=
∫
−
3
2
2
e
−
1
2
(
3
x
−
2
)
d
x
{\displaystyle \int -{\frac {3}{2}}{\sqrt {\frac {2}{e^{3x-2}}}}\,dx=\int -{\frac {3{\sqrt {2}}}{2}}e^{-{\frac {1}{2}}(3x-2)}\,dx}
From there, it becomes obvious to let
u
=
−
1
2
(
3
x
−
2
)
,
d
u
=
−
3
2
d
x
{\displaystyle u=-{\frac {1}{2}}(3x-2),\qquad du=-{\frac {3}{2}}\,dx}
.
Then,
∫
−
3
2
2
e
−
1
2
(
3
x
−
2
)
d
x
=
∫
2
e
u
d
u
=
2
e
u
+
C
=
2
e
−
1
2
(
3
x
−
2
)
+
C
=
2
e
3
x
−
2
+
C
{\displaystyle {\begin{aligned}\int -{\frac {3{\sqrt {2}}}{2}}e^{-{\frac {1}{2}}(3x-2)}\,dx&=\int {\sqrt {2}}e^{u}\,du\\&={\sqrt {2}}e^{u}+C\\&={\sqrt {2}}e^{-{\frac {1}{2}}(3x-2)}+C\\&=\mathbf {{\sqrt {\frac {2}{e^{3x-2}}}}+C} \end{aligned}}}
18.
∫
x
sec
(
x
2
−
5
)
tan
(
x
2
−
5
)
50
x
2
−
5
d
x
{\displaystyle \int {\frac {x\sec \left({\sqrt {x^{2}-5}}\right)\tan \left({\sqrt {x^{2}-5}}\right)}{50{\sqrt {x^{2}-5}}}}\,dx}
.
Let
u
=
x
2
−
5
,
d
u
=
x
x
2
−
5
d
x
{\displaystyle u={\sqrt {x^{2}-5}},\qquad du={\frac {x}{\sqrt {x^{2}-5}}}\,dx}
.
Then,
∫
x
sec
(
x
2
−
5
)
tan
(
x
2
−
5
)
50
x
2
−
5
d
x
=
∫
1
50
sec
(
u
)
tan
(
u
)
d
u
=
1
50
sec
(
u
)
+
C
=
sec
(
x
2
−
5
)
50
+
C
{\displaystyle {\begin{aligned}\int {\frac {x\sec \left({\sqrt {x^{2}-5}}\right)\tan \left({\sqrt {x^{2}-5}}\right)}{50{\sqrt {x^{2}-5}}}}\,dx&=\int {\frac {1}{50}}\sec(u)\tan(u)\,du\\&={\frac {1}{50}}\sec(u)+C\\&=\mathbf {{\frac {\sec \left({\sqrt {x^{2}-5}}\right)}{50}}+C} \end{aligned}}}
Let
u
=
x
2
−
5
,
d
u
=
x
x
2
−
5
d
x
{\displaystyle u={\sqrt {x^{2}-5}},\qquad du={\frac {x}{\sqrt {x^{2}-5}}}\,dx}
.
Then,
∫
x
sec
(
x
2
−
5
)
tan
(
x
2
−
5
)
50
x
2
−
5
d
x
=
∫
1
50
sec
(
u
)
tan
(
u
)
d
u
=
1
50
sec
(
u
)
+
C
=
sec
(
x
2
−
5
)
50
+
C
{\displaystyle {\begin{aligned}\int {\frac {x\sec \left({\sqrt {x^{2}-5}}\right)\tan \left({\sqrt {x^{2}-5}}\right)}{50{\sqrt {x^{2}-5}}}}\,dx&=\int {\frac {1}{50}}\sec(u)\tan(u)\,du\\&={\frac {1}{50}}\sec(u)+C\\&=\mathbf {{\frac {\sec \left({\sqrt {x^{2}-5}}\right)}{50}}+C} \end{aligned}}}
19.
∫
2
sec
2
(
ln
(
x
)
)
tan
(
ln
(
x
)
)
x
d
x
{\displaystyle \int {\frac {2\sec ^{2}\left(\ln(x)\right)\tan \left(\ln(x)\right)}{x}}\,dx}
.
Let
u
=
ln
(
x
)
,
d
u
=
1
x
d
x
{\displaystyle u=\ln(x),\qquad du={\frac {1}{x}}\,dx}
.
Then,
∫
2
sec
2
(
ln
(
x
)
)
tan
(
ln
(
x
)
)
x
d
x
=
∫
2
sec
2
(
u
)
tan
(
u
)
d
u
{\displaystyle \int {\frac {2\sec ^{2}\left(\ln(x)\right)\tan \left(\ln(x)\right)}{x}}\,dx=\int 2\sec ^{2}(u)\tan(u)\,du}
.
Let
v
=
tan
(
u
)
,
d
v
=
sec
2
(
u
)
d
u
{\displaystyle v=\tan(u),\qquad dv=\sec ^{2}(u)\,du}
.
Therefore,
∫
2
sec
2
(
u
)
tan
(
u
)
d
u
=
∫
2
v
d
v
=
v
2
+
C
=
tan
2
(
u
)
+
C
=
tan
2
(
ln
(
x
)
)
+
C
{\displaystyle {\begin{aligned}\int 2\sec ^{2}(u)\tan(u)\,du&=\int 2v\,dv\\&=v^{2}+C\\&=\tan ^{2}(u)+C\\&=\mathbf {\tan ^{2}\left(\ln(x)\right)+C} \end{aligned}}}
Alternatively, this could all be done with one substitution if one realized that
d
d
x
(
tan
[
ln
(
x
)
]
)
=
sec
2
(
ln
(
x
)
)
x
{\displaystyle {\frac {d}{dx}}\left(\tan \left[\ln(x)\right]\right)={\frac {\sec ^{2}\left(\ln(x)\right)}{x}}}
.Let
u
=
ln
(
x
)
,
d
u
=
1
x
d
x
{\displaystyle u=\ln(x),\qquad du={\frac {1}{x}}\,dx}
.
Then,
∫
2
sec
2
(
ln
(
x
)
)
tan
(
ln
(
x
)
)
x
d
x
=
∫
2
sec
2
(
u
)
tan
(
u
)
d
u
{\displaystyle \int {\frac {2\sec ^{2}\left(\ln(x)\right)\tan \left(\ln(x)\right)}{x}}\,dx=\int 2\sec ^{2}(u)\tan(u)\,du}
.
Let
v
=
tan
(
u
)
,
d
v
=
sec
2
(
u
)
d
u
{\displaystyle v=\tan(u),\qquad dv=\sec ^{2}(u)\,du}
.
Therefore,
∫
2
sec
2
(
u
)
tan
(
u
)
d
u
=
∫
2
v
d
v
=
v
2
+
C
=
tan
2
(
u
)
+
C
=
tan
2
(
ln
(
x
)
)
+
C
{\displaystyle {\begin{aligned}\int 2\sec ^{2}(u)\tan(u)\,du&=\int 2v\,dv\\&=v^{2}+C\\&=\tan ^{2}(u)+C\\&=\mathbf {\tan ^{2}\left(\ln(x)\right)+C} \end{aligned}}}
Alternatively, this could all be done with one substitution if one realized that
d
d
x
(
tan
[
ln
(
x
)
]
)
=
sec
2
(
ln
(
x
)
)
x
{\displaystyle {\frac {d}{dx}}\left(\tan \left[\ln(x)\right]\right)={\frac {\sec ^{2}\left(\ln(x)\right)}{x}}}
.
20.
∫
(
e
x
−
1
)
x
e
x
−
2
+
x
e
x
−
1
e
x
ln
(
x
)
d
x
{\displaystyle \int \left(e^{x}-1\right)x^{e^{x}-2}+x^{e^{x}-1}e^{x}\ln(x)\,dx}
.
It may be easier to see what to substitute once the integrand is written in an equivalent form.
∫
(
e
x
−
1
)
x
e
x
−
2
+
x
e
x
−
1
e
x
ln
(
x
)
d
x
=
∫
x
e
x
−
1
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
=
∫
e
(
e
x
−
1
)
ln
(
x
)
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
{\displaystyle {\begin{aligned}\int \left(e^{x}-1\right)x^{e^{x}-2}+x^{e^{x}-1}e^{x}\ln(x)\,dx&=\int x^{e^{x}-1}\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx\\&=\int e^{\left(e^{x}-1\right)\ln(x)}\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx\end{aligned}}}
From there, let
u
=
(
e
x
−
1
)
ln
(
x
)
,
d
u
=
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
{\displaystyle u=\left(e^{x}-1\right)\ln(x),\qquad du=\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx}
.
Then,
∫
e
(
e
x
−
1
)
ln
(
x
)
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
=
∫
e
u
d
u
=
e
u
+
C
=
e
(
e
x
−
1
)
ln
(
x
)
+
C
=
x
e
x
−
1
+
C
{\displaystyle {\begin{aligned}\int e^{\left(e^{x}-1\right)\ln(x)}\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx&=\int e^{u}\,du\\&=e^{u}+C\\&=e^{\left(e^{x}-1\right)\ln(x)}+C\\&=\mathbf {x^{e^{x}-1}+C} \end{aligned}}}
It may be easier to see what to substitute once the integrand is written in an equivalent form.
∫
(
e
x
−
1
)
x
e
x
−
2
+
x
e
x
−
1
e
x
ln
(
x
)
d
x
=
∫
x
e
x
−
1
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
=
∫
e
(
e
x
−
1
)
ln
(
x
)
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
{\displaystyle {\begin{aligned}\int \left(e^{x}-1\right)x^{e^{x}-2}+x^{e^{x}-1}e^{x}\ln(x)\,dx&=\int x^{e^{x}-1}\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx\\&=\int e^{\left(e^{x}-1\right)\ln(x)}\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx\end{aligned}}}
From there, let
u
=
(
e
x
−
1
)
ln
(
x
)
,
d
u
=
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
{\displaystyle u=\left(e^{x}-1\right)\ln(x),\qquad du=\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx}
.
Then,
∫
e
(
e
x
−
1
)
ln
(
x
)
(
e
x
−
1
x
+
e
x
ln
(
x
)
)
d
x
=
∫
e
u
d
u
=
e
u
+
C
=
e
(
e
x
−
1
)
ln
(
x
)
+
C
=
x
e
x
−
1
+
C
{\displaystyle {\begin{aligned}\int e^{\left(e^{x}-1\right)\ln(x)}\left({\frac {e^{x}-1}{x}}+e^{x}\ln(x)\right)\,dx&=\int e^{u}\,du\\&=e^{u}+C\\&=e^{\left(e^{x}-1\right)\ln(x)}+C\\&=\mathbf {x^{e^{x}-1}+C} \end{aligned}}}
21.
∫
2
sec
2
(
ln
(
x
2
+
2
x
)
)
x
3
−
1
x
4
+
2
x
d
x
{\displaystyle \int 2\sec ^{2}\left(\ln \left(x^{2}+{\frac {2}{x}}\right)\right){\frac {x^{3}-1}{x^{4}+2x}}\,dx}
.
Let
u
=
ln
(
x
2
+
2
x
)
,
d
u
=
2
x
3
−
1
x
4
+
2
x
d
x
d
x
{\displaystyle u=\ln \left(x^{2}+{\frac {2}{x}}\right),\qquad du=2{\frac {x^{3}-1}{x^{4}+2x}}\,dx\,dx}
.
Then,
∫
2
sec
2
(
ln
(
x
2
+
2
x
)
)
x
3
−
1
x
4
+
2
x
d
x
=
∫
sec
2
(
u
)
d
u
=
tan
(
u
)
+
C
=
tan
(
ln
(
x
2
+
2
x
)
)
+
C
{\displaystyle {\begin{aligned}\int 2\sec ^{2}\left(\ln \left(x^{2}+{\frac {2}{x}}\right)\right){\frac {x^{3}-1}{x^{4}+2x}}\,dx&=\int \sec ^{2}(u)\,du\\&=\tan(u)+C\\&=\mathbf {\tan \left(\ln \left(x^{2}+{\frac {2}{x}}\right)\right)+C} \end{aligned}}}
Let
u
=
ln
(
x
2
+
2
x
)
,
d
u
=
2
x
3
−
1
x
4
+
2
x
d
x
d
x
{\displaystyle u=\ln \left(x^{2}+{\frac {2}{x}}\right),\qquad du=2{\frac {x^{3}-1}{x^{4}+2x}}\,dx\,dx}
.
Then,
∫
2
sec
2
(
ln
(
x
2
+
2
x
)
)
x
3
−
1
x
4
+
2
x
d
x
=
∫
sec
2
(
u
)
d
u
=
tan
(
u
)
+
C
=
tan
(
ln
(
x
2
+
2
x
)
)
+
C
{\displaystyle {\begin{aligned}\int 2\sec ^{2}\left(\ln \left(x^{2}+{\frac {2}{x}}\right)\right){\frac {x^{3}-1}{x^{4}+2x}}\,dx&=\int \sec ^{2}(u)\,du\\&=\tan(u)+C\\&=\mathbf {\tan \left(\ln \left(x^{2}+{\frac {2}{x}}\right)\right)+C} \end{aligned}}}
30. Consider the integral
∫
sin
(
x
)
cos
(
x
)
d
x
{\displaystyle \int \sin(x)\cos(x)\,dx}
. Find the integral in two different ways. (a) Integrate by parts with
u
=
sin
(
x
)
{\displaystyle u=\sin(x)}
and
v
′
=
cos
(
x
)
{\displaystyle v'=\cos(x)}
. (b) Integrate by parts with
u
=
cos
(
x
)
{\displaystyle u=\cos(x)}
and
v
′
=
sin
(
x
)
{\displaystyle v'=\sin(x)}
. Compare your answers. Are they the same?
40.
∫
1
x
2
+
a
2
d
x
{\displaystyle \int {\frac {1}{x^{2}+a^{2}}}\,dx}
Let
x
=
a
tan
θ
;
d
x
=
a
sec
2
θ
d
θ
{\displaystyle x=a\tan \theta ;\qquad dx=a\sec ^{2}\theta d\theta }
Then
∫
1
x
2
+
a
2
d
x
=
∫
a
sec
2
θ
d
θ
a
2
(
tan
2
θ
+
1
)
=
∫
sec
2
θ
d
θ
a
sec
2
θ
=
1
a
∫
d
θ
=
θ
a
+
C
=
1
a
arctan
x
a
+
C
{\displaystyle {\begin{aligned}\int {\frac {1}{x^{2}+a^{2}}}dx&=\int {\frac {a\sec ^{2}\theta d\theta }{a^{2}(\tan ^{2}\theta +1)}}\\&=\int {\frac {\sec ^{2}\theta d\theta }{a\sec ^{2}\theta }}\\&={\frac {1}{a}}\int d\theta \\&={\frac {\theta }{a}}+C\\&=\mathbf {{\frac {1}{a}}\arctan {\frac {x}{a}}+C} \end{aligned}}}
Let
x
=
a
tan
θ
;
d
x
=
a
sec
2
θ
d
θ
{\displaystyle x=a\tan \theta ;\qquad dx=a\sec ^{2}\theta d\theta }
Then
∫
1
x
2
+
a
2
d
x
=
∫
a
sec
2
θ
d
θ
a
2
(
tan
2
θ
+
1
)
=
∫
sec
2
θ
d
θ
a
sec
2
θ
=
1
a
∫
d
θ
=
θ
a
+
C
=
1
a
arctan
x
a
+
C
{\displaystyle {\begin{aligned}\int {\frac {1}{x^{2}+a^{2}}}dx&=\int {\frac {a\sec ^{2}\theta d\theta }{a^{2}(\tan ^{2}\theta +1)}}\\&=\int {\frac {\sec ^{2}\theta d\theta }{a\sec ^{2}\theta }}\\&={\frac {1}{a}}\int d\theta \\&={\frac {\theta }{a}}+C\\&=\mathbf {{\frac {1}{a}}\arctan {\frac {x}{a}}+C} \end{aligned}}}