Integration Involving Linear Substitution
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Integrating a function with a linear substitution is the same as performing the chain rule in reverse. In Core four we formalize the knowledge that we learned in core three. Yet the procedure that we learn in core three is the same in core 4. In order to integrate the function
∫
f
′
[
g
(
x
)
]
.
g
′
(
x
)
d
x
=
f
[
g
(
x
)
]
+
c
{\displaystyle \int f^{'}[g(x)].g^{'}(x)\,dx=f[g(x)]+c}
Integrate 12cos(12x + 9).
g(x) = 12x + 9 and
d
g
(
x
)
d
x
=
12
{\displaystyle {\frac {dg(x)}{dx}}=12}
sin(12x+9) + C Integration by Parts
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Integration by parts does not actually solve the problem as the method of substitution does, instead it changes the function into something that is (hopefully) easier to integrate, either by substitution or otherwise.
Consider the product rule:
d
y
d
x
=
v
d
u
d
x
+
u
d
v
d
x
{\displaystyle {\frac {dy}{dx}}=v{\frac {du}{dx}}+u{\frac {dv}{dx}}}
Integrating this rule gives:
∫
d
y
d
x
d
x
=
∫
v
d
u
d
x
d
x
+
∫
u
d
v
d
x
d
x
{\displaystyle \int {\frac {dy}{dx}}\,dx=\int v{\frac {du}{dx}}\,dx+\int u{\frac {dv}{dx}}\,dx}
⇒
∫
d
y
d
x
d
x
−
∫
v
d
u
d
x
d
x
=
∫
u
d
v
d
x
d
x
{\displaystyle \Rightarrow \int {\frac {dy}{dx}}\,dx-\int v{\frac {du}{dx}}\,dx=\int u{\frac {dv}{dx}}\,dx}
⇒
∫
u
d
v
d
x
d
x
=
u
v
−
∫
v
d
u
d
x
d
x
{\displaystyle \Rightarrow \int u{\frac {dv}{dx}}\,dx=uv-\int v{\frac {du}{dx}}\,dx}
The function we need to integrate has now changed from
u
d
v
d
x
{\displaystyle u{\tfrac {dv}{dx}}}
v
d
u
d
x
{\displaystyle v{\tfrac {du}{dx}}}
Integrate
x
e
x
{\displaystyle xe^{x}}
x
{\displaystyle x}
u
=
x
⇒
d
u
d
x
=
1
{\displaystyle u=x\Rightarrow {\tfrac {du}{dx}}=1}
d
v
d
x
=
e
x
⇒
v
=
e
x
{\displaystyle {\tfrac {dv}{dx}}=e^{x}\Rightarrow v=e^{x}}
∫
x
e
x
d
x
=
x
e
x
−
∫
1
⋅
e
x
d
x
{\displaystyle \int xe^{x}\,dx=xe^{x}-\int 1\cdot e^{x}\,dx}
=
x
e
x
−
e
x
+
c
{\displaystyle =xe^{x}-e^{x}+c\,\!}
=
(
x
−
1
)
e
x
+
c
{\displaystyle =(x-1)e^{x}+c\,\!}
Another more complex example:
Integrate
x
2
e
x
{\displaystyle x^{2}e^{x}}
x
{\displaystyle x}
u
=
x
2
⇒
d
u
d
x
=
2
x
{\displaystyle u=x^{2}\Rightarrow {\frac {du}{dx}}=2x}
d
v
d
x
=
e
x
⇒
v
=
e
x
{\displaystyle {\frac {dv}{dx}}=e^{x}\Rightarrow v=e^{x}}
∫
x
2
e
x
d
x
=
x
2
e
x
−
∫
2
x
⋅
e
x
d
x
{\displaystyle \int x^{2}e^{x}\,dx=x^{2}e^{x}-\int 2x\cdot e^{x}\,dx}
Now we can use integration by parts again on
∫
2
x
⋅
e
x
d
x
{\displaystyle \int 2x\cdot e^{x}\,dx}
u
=
2
x
⇒
d
u
d
x
=
2
{\displaystyle u=2x\Rightarrow {\frac {du}{dx}}=2}
d
v
d
x
=
e
x
⇒
v
=
e
x
{\displaystyle {\frac {dv}{dx}}=e^{x}\Rightarrow v=e^{x}}
x
2
e
x
−
∫
2
x
⋅
e
x
d
x
=
x
2
e
x
−
(
2
x
e
x
−
∫
2
e
x
d
x
)
{\displaystyle x^{2}e^{x}-\int 2x\cdot e^{x}\,dx=x^{2}e^{x}-(2xe^{x}-\int 2e^{x}\,dx)}
=
x
2
e
x
−
2
x
e
x
+
2
e
x
+
c
{\displaystyle =x^{2}e^{x}-2xe^{x}+2e^{x}+c\,\!}
=
e
x
(
x
2
−
2
x
+
2
)
+
c
{\displaystyle =e^{x}(x^{2}-2x+2)+c\,\!}
![{\displaystyle \int f^{'}[g(x)].g^{'}(x)\,dx=f[g(x)]+c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ae7f26f064c993232fdfbaf7b92fa54fbf983a4)
