Integration by Parts and U-Substitution

Our aim in this video is to find the antiderivative of the function square root of x + 1 times cosine of square root of x + 1
This is a typical calculus 2 problem.
We will carefully walk you through the solution.
When we look at this problem, it is clear that this is a complex integration question.
Whenever you are confronted with this type of problems, it is important to devise a strategy to attack them efficiently.
It is always beneficial to analyze the function that we are asked to integrate. This initial analysis of the function will help us figure out the proper strategy to solve the problem.
In this case, the function square root of x + 1 times cosine of square root of x + 1 is a product of a radical function square root of x + 1 and a composition of two functions: the cosine function and the square root function.
Because this product involves a radical function and a composition of a nonlinear function with the same radical function, it is a good idea to start with the method of substitution.
We will perform a change of variable, we introduce a new variable.
It is evident that the natural choice for the new variable y should be the inner function. Thus, y should be square root of x + 1 .
So, we set y equal to square root of x + 1 and d y should be equal to the fraction 1 , over 2 times square root of x + 1 .
The integral of square root of x + 1 times cosine of square root of x + 1 d x , can be expressed as the integral of 2 times, square root of x + 1 times, square root of x + 1 times cosine of square root of x + 1 , times the fraction 1 , over 2 times square root of x + 1 d x .
We can clearly see y and d y in the expression of the integral.
So, we replace square root of x + 1 by y, and the fraction 1 , over 2 times square root of x + 1 by d y. We then obtain the new expression of this integral as the integral of 2 y square, times, cosine of y d y.
This new expression of the function to integrate is simpler than the original one, however we still have to carefully figure out the best way to integrate this new function.
This new function is the product of a polynomial function, 2 times u square, and the nonlinear function, cosine of y .
remark that , the best method for the integration of products of polynomial functions and nonlinear transcendental functions is the method of integration by parts.
In this case, we have to use the integration by parts method to compute the integral of 2 y square times, the function cosine of y d y.
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Let us integrate the function 2 y square times cosine of y .
note that , at first glance, we easily see that the function we have to integrate is a product of a nonlinear function, cosine of y , with a polynomial function 2 y square .
occasionally, 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
remark that , The polynomial function 2 y square is of degree 2.