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Integration by Substitution

The studying process about the integration by substitution method involves the solving process of the integral function. If we have the function that are not activate or convenient to integrate directly are solved by different methods of the integration. Integration by substitution, also known as u-substitution or change of variables.

Integration by Substitution Method:
Let A : [c, d] -> R be continuously differentiable. Assume A[c, d] = [m, n] with A(c) = m and A(d) = n. If g : [m, n] -> R is continuous, then g(A)A' is Riemann integrable on [c, d] and $\int_c^d$ g(A(x))A'(x) dx = $\int_m^n$g

Related Calculators
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Integration by Substitution Examples

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Below are the examples on integration by substitution -
Example 1 : Integrate the function  `int 10/sqrt(1 - 100x^2) dx`.


 Let I = `int 10/sqrt(1 - 100x^2) dx`              

 The step is to be involved for substitution by splitting the term `10/sqrt(1- 100x^2)`

Put 10 x = sin u                                    … (1)

       `=>` u = sin?1 (10x)

and 10 dx = cos u du                                   … (2)

? I =  `int 10/sqrt(1- 100x^2) dx`      

=`int 1/sqrt(1 - sin^2 u) (cos u du)`    by using (1) and (2)

= `int 1/sqrt(cos^2 u) (cos u du)`   

= `int` du
= u + c                 (Put u = sin?1(10x) in the function for the variable we assigned)

=>  `int 10/sqrt(1 - 100x^2) dx` = sin?1 (10x) + c, where c is the constant of integration.      

 Example 2 :   Integrate the function  `int 5/(1 + x^2) dx` using the substitution methods.


Let I = `int 5/(1 + x^2) dx`            

Put x = tan u                                   … (1)

    `=>`  u = tan?1 x

and dx = sec2 u du                                   … (2)

? I =  `int 5/(1+ tan^2u) sec^2u du`          by using (1) and (2)  

    =   `int 5/(sec^2u) sec^2u du`  

    = `int` 5du
    = 5u + c                 (Put  u = tan?1 x in the function for the variable we assigned)

  => `int 5/(1+ x^2) dx` =  5 tan?1 x  + c
This is the solution obtained after substituting the terms.

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