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Revision:Integration by Parts

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Integration by Parts

From the product rule, we can obtain the following formula, which is very useful in integration:

\displaystyle \int^b_a u\frac{dv}{dx}dx=[uv]^b_a - \int^b_a v\frac{du}{dx}dx

It is used when integrating the product of two functions (u and \frac{dv}{dx}).


Contents

Deriving the Formula

The integration by parts formula is derived from the product rule, used in differentiation.

If u and v are functions of x,

\frac{d}{dx}(uv)=u\frac{dv}{dx}+ v\frac{du}{dx}

Integrating each side with respect to x:

uv=\displaystyle \int u\frac{dv}{dx} \ dx + \int v\frac{du}{dx} \ dx

Rearranging:

\displaystyle \int u\frac{dv}{dx} \ dx=uv - \int v\frac{du}{dx} \ dx


Integrating ln x

Integration by parts provides us with the method for integrating \ln x, but it is not obvious at first. Here is the worked example.

I = \displaystyle \int \ln x \ dx

Let u=\ln x

\Rightarrow \frac{du}{dx}=\frac{1}{x}

Let \frac{dv}{dx}=1

\Rightarrow v=x

Putting it into the formula:

I=x\ln x-\displaystyle \int 1dx

\Rightarrow I=x\ln x-x+c


Uses in A-Level Exams

The technique of integration by parts will often by indicated by explicitly stating so in the question. However, there is a chance you will have to decide for yourself, in which case it is useful to have an idea of when to use the method.

An example of when to use integration by parts is when we have a x term, multiplied by a trigonometric or exponential function. Here is an example.

I = \displaystyle \int_0^\pi x\sin x \ dx

Let u=x

\Rightarrow \frac{du}{dx}=1

Let \frac{dv}{dx}=\sin x

\Rightarrow v=-\cos x

Putting it into the formula: I=[x(-\cos x)]_0^\pi - \displaystyle \int_0^\pi (-\cos x)1 \ dx

I=\pi(-\cos\pi)-0+\displaystyle \int_0^\pi \cos x \ dx

I=\pi+[\sin x]_0^\pi

I=\pi+\sin\pi-\sin0

I=\pi


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