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mathsIntegral CalculusVarious Forms and Results of Integrals

### Integration by Parts

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Integration of functions that are product of sub-functions is hard to find result for. For example, f(x)=x^2 and g(x)=sinx can be individually integrated. But integrating f(x)g(x) = x^2 sinx is to be worked out.

Knowing that integration is anti-derivative, let us examine how product of sub-functions is differentiated.
(uv)′ = vu′+uv′

If one of the functions is in x^n form, then repeated differentiation will result in a constant. Knowing this, the equation above is modified to
u v′ = (uv)′ - vu′

integrating this
int u v′ dx = int (uv)′ dx - int v u′ dx

int u v′ dx = uv - int v u′ dx

substituting u= f(x) and v′ = g(x) and so v= int g(x) dx

int f(x) g(x) dx  = f(x) intg(x)dx - int [int g(x) dx] [d/(dx)f(x)] dx

The advantage is, the function f(x) is differentiated in the second term. If f(x)=x then the second term is simplified to a standard result.

int f(x) g(x) dx  = f(x) intg(x)dx - int [int g(x) dx] [d/(dx)f(x)] dx

This is called integration by parts. Note that the integrand is split into two parts.

int x^2 sin x dx

using integration by parts
=-x^2cos x - int (-cos x) (2x)dx+c

=-x^2cos x + int 2xcos x dx+c

How to proceed further in integration?

• The integration by parts can be used on the second term too
• The integration by parts can be used on the second term too
• no further steps are possible to integrate

The answer is "The integration by parts can be used on the second term too"

=-x^2cos x + int 2xcos x dx+c

using integration by parts on the second term
=-x^2cos x + 2x sin x - int sinx xx2 dx+ c

= -x^2cos x + 2x sin x + 2cos x+ c

Interesting observation in the integration by parts is that one of the two sub-functions is differentiated sequentially. The other is integrated repeatedly. If one of the functions is e^x, then as the sequential integration or differentiation does not change e^x. This leads to an interesting form of integration.

int e^x sin x dx

= e^x(-cos x)+ int e^x cos x dx + c

= -e^xcos x + e^x sinx - int e^x sin x dx + c

The result has the integral we started off with. How to resolve this?

• handle this as an equation and rearrange such that the term is in one side of the equation
• handle this as an equation and rearrange such that the term is in one side of the equation
• no further steps are possible to integrate

The answer is "handle this as an equation and rearrange such that the term is in one side of the equation".

I = int e^x sin x dx

= e^x(-cos x)+ int e^x cos x dx + c

= -e^xcos x + e^x sinx - int e^x sin x dx + c

The above is written as I = -e^xcos x + e^x sinx - I + c

what is the result I?

• I=(-e^xcos x + e^x sinx)//2 + c
• I=(-e^xcos x + e^x sinx)//2 + c
• I is canceled on both the sides

The answer is "I=(-e^xcos x + e^x sinx)//2 + c".

Another interesting observation in integration by parts: If one of the functions is e^x, then as the sequential integration or differentiation does not change e^x. This leads to an interesting form of integration when the other function is in the form. h(x)+h′(x).

int e^x (h(x)+h′(x)) dx

= int e^x h(x) dx + int e^x h′(x) dx

applying integration by parts in the first integral

= e^x h(x) - int e^x h′(x) dx + int e^x 1/x dx +c

Is it possible to simplify this further?

• yes, cancel the second term and third term
• yes, cancel the second term and third term
• no further steps are possible to integrate

The answer is "yes, cancel the second term and third term". The integration result is e^xh(x) + c.

Integrate int e^x (ln x + 1/x) dx.

int e^x (ln x + 1/x) dx

= int e^x ln x dx + int e^x 1/x dx  = e^x ln x - int e^x 1/x dx + int e^x 1/x dx +c = e^x ln x +c

Integration by Parts: Integrand is taken as product of two functions to integrate.

Integration by Parts: int f(x) g(x) dx  = f(x) int g(x)dx - int [int g(x) dx] [d/(dx)f(x)] dx

e^x sin x or e^x cos x in Integration by Parts: After two iterations of integration by parts, the integral is solved by rearranging the terms of the equation.

e^x in Integration by Parts: int e^x (h(x)+h′(x)) dx =e^x h(x) + c

Solved Exercise Problem:

Integrate int e^x arctan x + e^x/(1+x^2) dx

• e^x arctan x + c
• e^x arctan x + c
• e^x/(1+x^2) + c

The answer is "e^x arctan x + c". The integral is in the form int e^x (h(x)+h′(x)) dx

Solved Exercise Problem:

Integrate int e^x cos x dx

• e^x (sinx -cos x)//2 + c
• e^x (sin x+cos x)//2 + c
• e^x (sin x+cos x)//2 + c

The answer is "e^x (sin x+cos x)//2 + c".

Solved Exercise Problem:

Integrate int x sin x dx

• -x cos x + sinx +c
• -x cos x + sinx +c
• x cos x + sin x + c

The answer is "-x cos x + sinx +c".

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