Welcome to Live Chat

Welcome to LiveWebTutors Services, World's leading Academic solutions provider with Millions of Happy Students.

Call Back

24x7 Support Available

To Get the Best Price Chat With Our Experts

chat now

In A Hurry? Get A Callback

Table of Content

Integration by Parts

The mathematicsmethod of integration by parts is very useful method in the integral of a product of two or more functions. This method is very helpful in comparing with another method of integration. Let we have two functions like f (x) and g (x) in product form then we have the formula for integration by parts: - Integration of f (x) * g (x) = f (x) * integration of g (x) – integration of [f’(x) * integration of g(x)]. In other words the integral of the product of two functions = first functions * integration of second function – integral of (differentiation of first function * integral of second function). Here is the most important part to choose first and second function.

If we have two functions in product form if one function of these two functions is not directly integrable (like sin^-1 x, cos^-1 x, tan^-1 x, cot^-1 x, sec^-1 x, csc^-1 x, log x etc) these functions we take as a first part. If in product of two functions both functions are easily integrable, then we choose first functions as the derivative of this function is a simple function. Thus we can easily integrate the given function. Always product of two functions, in order to choose the first function, first we consider the inverse trigonometric function, logarithmic function, algebraic function, trigonometric function, exponential functions. This is the order to choose the first function.

Let us see some examples of integration by parts. Example: - Integrate the given function by using integration by parts x^3 cos x. Solution: - given function in product form ( x^2 + x + 1) cos x. Here we choose first function algebraic function f (x) = x^2 + x + 1 and letsthe second function is g(x) cos x. Now using formula integration by parts: - the integral of the product of two functions = first functions * integration of second function – integral of (differentiation of first function * integral of second function).

Integral of [(x^2 + x + 1) cos x] = (x^2 + x + 1) sin x – integral of [(2x + 1)sinx], again using integration by parts. And get integral of [(2x + 1)sinx] = - (2x + 1) cos x – integral of [(2) (-cos x) = -(2x + 1) cos x + 2 sin x, Now we plug the value of the integral of [(2x + 1)sinx] and get Integral of [(x^2 + x + 1) cos x] = (x^2 + x + 1) sin x – [-(2x + 1) cos x + 2 sin x], è Integral of [(x^2 + x + 1) cos x] = (x^2 + x + 1) sin x + (2x + 1) cos x - 2 sin x]. In this type we can integrate functions. This is how we can find the integral of many functions using integration by parts method. This method helps in finding integral of those functions which cannot be solved using the normal method.

Our Amazing Features
  • On Time Delivery

    There is no deadline that can stop our writers from delivering quality assignments to the students.

  • Plagiarism Free Work

    Get authentic and unique assignments by using our 100% plagiarism-free services.

  • 24 X 7 Live Help

    The experienced team of Live web tutors has got your back at all times of the day. Get connected with our customer support executives to receive instant and live solutions for your assignment problems.

  • Services For All Subjects

    We can build quality assignments in the subjects you're passionate about. Be it Engineering and Literature or Law and Marketing we have an expert writer for all.

  • Best Price Guarantee

    Get premium service at a pocket-friendly rate. At live web tutors, we understand the tight budget of students and thus offer our services at highly affordable prices.

live review Our Mission Client Satisfaction
  • I ordered my HRM assignment from this platform. The solution helped me to pass my assessment with flying colors. Well done! Keep up the good work.

    27 Oct 2020


  • Recently, I ordered my history case study from this platform. The professionals did great work with the content. I am yet to be graded for it but as far as I reviewed the text I am definitely impressed with the quality provided to me at this price. Thank you for your support.

    27 Oct 2020


  • I got good grades with my nursing assignment, all thanks to the experts of this platform. My writing skills are not that great hence the experts of this portal helped me to pass my program with good grades.

    27 Oct 2020


  • I ordered my English assignment from this platform. The service was very quick and impressive. The customer support executives were very polite and helpful. I am grateful for their assistance.

    27 Oct 2020


  • Overall, it was a pleasant experience. I ordered my management thesis from this platform. The experts did thorough and detailed work on my assignment. All my instructions were implemented precisely.

    27 Oct 2020


View All Review