## Formula for Integration by Parts

If u and v are two functions of x, then the formula for integration by parts is –

\(\int\) u.v dx = u \(\int\) v dx – \(\int\)[\(du\over dx\).\(\int\)v dx]dx

i.e The integral of the product of two functions = (first function) \(\times\) (Integral of Second function) – Integral of { (Diff. of first function) \(\times\) (Integral of Second function)}

**Note** – We can choose the first function as the function which comes first in the word **ILATE, **where

I – Stands for the Inverse Trigonometric Function

L – Stands for the Logarithmic Function

A – Stands for the Algebraic Function

T – Stands for the Trigonometric Function

E – Stands for the Exponential Function

Example : Solve the integral \(\int\) \(x sin3x\) dx using the formula for integration by parts.

Solution : Here both the functions viz. x and sin3x are easily integrable and the derivative of x is one, a less complicated function. Therefore, we take x as the first function and sin3x as the second function.

\(\therefore\) I = \(\int\) x cos3x dx

= x [\(\int\) sin3x dx] – \(\int\)[\({d\over dx}(x)\) \(\times\) \(\int\)sin3x dx] dx

= x (\(-1\over 3\) cos3x) – \(\int\) [\(-1\over 3\) cos3x] dx

\(\implies\) I = \(-1\over 3\) xcos3x + \(1\over 3\) \(\int\) cos3x dx

\(\implies\) I = -\(1\over 3\) xcos3x + \(1\over 9\) sin3x + C

Example : Solve the integral \(\int\) \(x sin^{-1}x\) dx using the formula for integration by parts.

Solution : Taking \(sin^{-1}x\) as the first function and x as the second function by using the ILATE rule.

I = \(\int\) \(x sin^{-1}x\)

= (\(sin^{-1}x\))\(x^2\over 2\) – \(\int\)[\(1\over \sqrt{1-x^2}\) \(\times\) \(x^2\over 2\)] dx

= \(x^2\over 2\)(\(sin^{-1}x\)) + \(1\over 2\) \(\int\)[\((-x)^2\over \sqrt{1-x^2}\) dx = \(x^2\over 2\)(\(sin^{-1}x\)) + \(1\over 2\) \(\int\)[\({1-x^2-1}\over \sqrt{1-x^2}\) dx

\(\implies\) I = \(x^2\over 2\)(\(sin^{-1}x\)) + \(1\over 2\) {\(\int\)[\({1-x^2}\over \sqrt{1-x^2}\) dx – \(1\over \sqrt{1-x^2}\) dx}

\(\implies\) I = \(x^2\over 2\)(\(sin^{-1}x\)) + \(1\over 2\) {\(\int\)[\(\sqrt{1-x^2}\) dx – \(1\over \sqrt{1-x^2}\) dx}

I = \(x^2\over 2\)(\(sin^{-1}x\)) + \(1\over 2\) {\({1\over 2}x\)[\(\sqrt{1-x^2}\) dx + \(1\over 2\) \(\sin^{-1}x\) – \(sin^{-1}x\)] + C

\(\implies\) I = \(x^2\over 2\)(\(sin^{-1}x\)) + \(1\over 4\)\(x \sqrt{1-x^2}\) dx – \(1\over 4\) \(\sin^{-1}x\) + C

Hope you learnt what is the formula for integration by parts, learn more concepts of Indefinite Integration and practice more questions to get ahead in the competition. Good luck!