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Integration by parts




Integration by Parts

If u and v are any two differentiable functions of a single variable x (say). Then, by
the product rule of differentiation, we have
$\frac {d}{dx} uv = u \frac {dv}{dx} + v \frac {du}{dx}$
Integrating both sides
$uv = \int u \frac {dv}{dx} + \int v \frac {dv}{dx}$
or
$\int u \frac {dv}{dx}= uv - \int v \frac {dv}{dx}$
if u =f(x) and $\frac {dv}{dx} = g(x)$
then
$\int f(x) g(x) = f(x)  (\int g(x) dx )- \int \left \{ \frac {df(x)}{dx} \int g(x) dx \right \}dx$
The above formula is very useful formula in doing the integrals
We can decide first function using the word ILATE
I -> Inverse trigonometric functions
L -> Logarithmic functions
A-> Algebraic functions
T -> trigonometric functions
E -> Exponential functions

Solved Examples

Example 1
\[ \int x \cos(x) \, dx \]
Solution
Choose:
\[ f(x) = x \]
\[ g(x) = \cos(x) \]
Using the formula:
\[ \int x \cos(x) \, dx = x\sin(x) - \int \sin(x) \, dx \]
\[ = x\sin(x) + \cos(x) + C \]

Example 2
\[ \int x^2 \ln(x) \, dx \]
Solution
Choose:
\[ f(x) = \ln(x) \]
\[ g(x) = x^2 \]
Using the formula:
\[ \int x^2 \ln(x) \, dx = \frac{1}{3}x^3 \ln(x) - \int \frac{1}{3}x^3 \cdot \frac{1}{x} \, dx \]
\[ = \frac{1}{3}x^3 \ln(x) - \frac{1}{3} \int x^2 \, dx \]
\[ = \frac{1}{3}x^3 \ln(x) - \frac{1}{9}x^3 + C \]

Example 3
\[ \int e^x \sin(x) \, dx \]
Solution
This is a unique example where applying integration by parts twice gets us back to our original integral. Let's see how:
Choose:
\[ f(x) = e^x \]
\[ g(x) = \sin(x) \]
Using the formula:
\[ \int e^x \sin(x) \, dx = -e^x\cos(x) - \int (-e^x\cos(x)) \, dx \]
Now, integrate \( -e^x\cos(x) \) using integration by parts again:
Choose:
\[ f(x) = e^x \]
\[ g(x) = \cos(x) \]
This gives:
\[ \int e^x \cos(x) \, dx = e^x\sin(x) - \int e^x \sin(x) \, dx \]

Substituting this result back into our original equation, we can solve for the integral. The final result is:
\[ \int e^x \sin(x) \, dx = \frac{e^x (\sin(x) - \cos(x))}{2} + C \]

Special Integrals based on Integration by Parts

A. $\int e^x{ f(x) + f^{'} (x) } dx =  e^x f(x)  + C$
Proof
$\int e^x{ f(x) + f^{'} (x) } dx= \int e^x f(x) dx + \int e^x f^{'} (x) dx$
Now lets calculate $\int e^x f(x) dx$ using integration by parts by taking f(x) and $e^x$ as the first function and second function
$\int e^x f(x) dx = f(x) e^x - \int e^x f^{'} (x) dx$
Substituting in Above we get
$\int e^x{ f(x) + f^{'} (x) } dx=f(x) e^x - \int e^x f^{'} (x) dx + \int e^x f^{'} (x) dx=e^x f(x)  + C $

B. $ \int \sqrt {a^2 - x^2} dx = \frac {1}{2} x \sqrt {a^2 - x^2} + \frac {1}{2} a^2 \sin^{-1} \frac {x}{a} + C$
$ \int \sqrt {a^2 + x^2} dx = \frac {1}{2} x \sqrt {a^2 + x^2} + \frac {1}{2} a^2  ln |x +\sqrt {a^2 + x^2}|  + C$
$ \int \sqrt {x^2 -a ^2} dx = \frac {1}{2} x \sqrt {x^2 - a^2} - \frac {1}{2} a^2  ln |x +\sqrt {x^2 - a^2}|  + C$
We can derive these formula either by trigonometric substitution or integration by parts
The above formula can be to use to integrate the below type of function
$ \int \sqrt {ax^2 + bx + c} dx$
We can convert $ax^2 + bx + c$ into above using square method

Solved examples

Example 1
\[ \int \sqrt {x^2 - 1} \, dx \]
Solution
we can use the below formula here with a =1
$ \int \sqrt {x^2 -a^2} dx = \frac {1}{2} x \sqrt {x^2 - a^2} - \frac {1}{2} a^2  ln |x +\sqrt {x^2 - a^2}|  + C$
Therefore
$ \int \sqrt {x^2 - 1} \, dx = \frac {1}{2} x \sqrt {x^2 - 1} - \frac {1}{2}   ln |x +\sqrt {x^2 - 1}|  + C$

Example 2
\[ \int \sqrt { 4 -x^2} \, dx \]
Solution
We can use the below formula here with a=2
$ \int \sqrt {a^2 - x^2} dx = \frac {1}{2} x \sqrt {a^2 - x^2} + \frac {1}{2} a^2 \sin^{-1} \frac {x}{a} + C$
Therefore
$\int \sqrt { 4 -x^2} \, dx=\frac {1}{2} x \sqrt {4 - x^2} + 2 \sin^{-1} \frac {x}{2} + C$

Example 3
\[ \int \sqrt { 3 -2x-x^2} \, dx \]
Solution
The above integral can be written as
\[ \int \sqrt { 3 -2x-x^2} \, dx \]
=\[ \int \sqrt {4 -(x+1)^2} \, dx \]
Now y=x+1
dy=dx
Therefore
$=\int \sqrt {4 -y^2} \, dy$
No wee can use the below formula here with a=2
$ \int \sqrt {a^2 - x^2} dx = \frac {1}{2} x \sqrt {a^2 - x^2} + \frac {1}{2} a^2 \sin^{-1} \frac {x}{a} + C$
So
$=\frac {1}{2} y \sqrt {4 - y^2} + 2 \sin^{-1} \frac {y}{2} + C$

Substituting back the value
$=\frac {1}{2} (x+1) \sqrt { 3 -2x-x^2} + 2 \sin^{-1} \frac {x+1}{2} + C$


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