Integration by Parts | Calculus 2

Our goal in Problem 1 is to find the antiderivative of the function 3 x square , , , + 4 x times cosine of 9 x .
Our aim in this problem is to find the antiderivative of the function 3 x square , , , + 4 x times cosine of 9 x .
Let us integrate the function 3 x square , , , + 4 x times cosine of 9 x .
it comes that , at first glance we easily see that the function we have to integrate is a product of a nonlinear function, cosine of 9 x , with a polynomial function 3 x square , , , + 4 x .
often , when you are faced with this type of complex integration problem, you have to adopt the appropriate strategy that will help you simplify the problem, and find the antiderivative efficiently.
The two main techniques that we learn in calculus 2, for solving integrals that involve the products or quotients of nonlinear functions are the U Substitution method and the method of integration by parts.
The crucial first step of our strategy is to decide which of these two classical methods we should use.
In this case, we have a product of a nonlinear function with a polynomial function, so the best method to use is the method of integration by parts. Because this method will help us reduce the complexity of the problem by reducing the degree of the polynomial
as a matter of fact, The polynomial function 3 x square , , , + 4 x is of degree 2.
Thus, , our first goal will be to reduce the degree of the polynomial function to 1, by using the method of integration by parts.
Let us briefly review the method of Integration by parts.
We know that the integral of u d v is equal to. u times v, minus the integral of v d u.
In sum, , we set u equal to 3 x square , , , + 4 x , so d u equals 6 x , , , + 4 d x.
And, d v equals to cosine of 9 x d x, and v equals to { 1 } over { 9 } times sine of 9 x .
lastly,, The integral of 3 x square , , , + 4 x times cosine of 9 x d x is equal to.
{ 3 x square , , , + 4 x } over { 9 } times sine of 9 x , plus the integral of { , minus 6 x , minus 4 } over { 9 } times sine of 9 x dx. Consequently we still have to integrate { , minus 6 x , minus 4 } over { 9 } times sine of 9 x , using integration by parts.
Let us integrate the function { , minus 6 x , minus 4 } over { 9 } times sine of 9 x .
Given that the function to integrate { , minus 6 x , minus 4 } over { 9 } times sine of 9 x is still the product of a nonlinear function with a polynomial function,
we still have to use the method of integration by parts.
The main aim of these repeated utilization of the method of integration by parts is to reduced the degree of the polynomial function { , minus 6 x , minus 4 } over { 9 }.
Thus, , we set u equal to { , minus 6 x , minus 4 } over { 9 }, so d u equals , minus 2 over 3 d x.
And, d v equals to sine of 9 x d x, and v equals to { , minus 1 } over { 9 } times cosine of 9 x .