# What is the Formula for Integration by Parts ?

## 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!