Integration by Parts Revision Notes for AQA A-Level Mathematics

8.2.8 Integration by Parts

Integration By Parts

When integrating a product of two (unrelated by differentiation) functions, we use integration by parts.

\text{e.g. } \int 3x^2 (7x^3 + 4x) \, dx

Since the outside is the differential of the inside ( $\times k$ ), these functions are "related by differentiation." In this instance, substitution is appropriate.

\text{e.g. } \int x \sin x \, dx

\xcancel{\int x\ xx\ dx = \frac {x^2}{2} \times \frac {x^2}{2}+c}

The two functions are not "related by differentiation," so integration by parts is appropriate.

Integration By Parts: Deriving the Formula

Note

Starting with the product rule for differentiation:

(uv)' = uv' + vu'

Swapped LHS and RHS:

uv' = (uv)' - vu'

\int uv' \, dx = \int (uv)' \, dx - \int vu' \, dx

Integrating both sides dx:

\int uv' \, dx = uv - \int vu' \, dx

When integrating the product of two functions, it can sometimes be simpler to use this formula:

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

Note

$\int x \sin x \, dx$ cannot be integrated in its current form. Using integration by parts:

Let $u = x \Rightarrow \frac{du}{dx} = 1$

v = \cos x \Leftarrow \frac{dv}{dx} = \sin x

\int x \sin x \, dx = uv - \int v \frac{du}{dx} \, dx = -x \cos x - \int -\cos x \times 1\, dx

= -x \cos x + \int \cos \, dx = = -x \cos x + \sin x + c

= -x \cos x + \sin x + c

Note

$\int xe^x \, dx$ Let $u = x \Rightarrow \frac{du}{dx} = 1$

v = e^x \Leftarrow \frac{dv}{dx} = e^x

\Rightarrow \int xe^x = xe^x - \int e^x \, dx = xe^x - e^x + C

Note

$\int x \ln(x) \, dx$ Cannot find $\int \ln(x) \, dx$ directly. Try $u=\ln(x)$ instead

Let $u = \ln(x) \Rightarrow \frac{du}{dx} = x^{-1}$

$v = \frac{x^2}{2} \Leftarrow \frac{dv}{dx} = x$

\Rightarrow \int x \ln(x) \, dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} x^{-1} \, dx

= \frac{x^2}{2}\ln x - \int \frac {x}{2}dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + c

Note

Key Point: $\ln(x)$ can't be directly integrated, so it cannot be the $\frac{dv}{dx}$ in integration by parts.

Note

\int \ln x \, dx \equiv \int 1 \times \ln x \, dx

\text {Let} \quad u = \ln x \Rightarrow \frac{du}{dx} = x^{-1}

\text {Let} \quad v = x \Leftarrow \frac{dv}{dx} = 1

\Rightarrow \int \ln x = x \ln x - \int x x^{-1}\, dx = x \ln x-\int 1 \,dx= x \ln x - x + c

Note

\int x^2 \cos x \, dx

\text {Let} \quad u = x^2 \Rightarrow \frac{du}{dx} = 2x

\text {Let} \quad v = \sin x \Rightarrow \frac{dv}{dx} = \cos x

\Rightarrow \int x^2 \cos x = x^2 \sin x - \int 2x \sin x \, dx \quad (*)

= x^2 \sin x - (-2x \cos x + 2 \sin x) + c

= x^2 \sin x + 2x \cos x - 2 \sin x + c

(*)\quad u = 2x \Rightarrow \frac {du}{dx} = 2 \quad \bigg|\quad v = -\cos x \Leftarrow \frac {dv}{dx} = \sin x

\int 2x \sin x \, dx = -2x \cos x - \int -2 \cos x \, dx

\int 2x \sin x \, dx = -2x \cos x + \int 2 \cos x \, dx

= -2x \cos x + 2 \sin x

Note

\int_1^2 x e^x \, dx \quad \text{Calculate the :highlight[exact value].}

\text {Let} \quad u = x \Rightarrow \frac{du}{dx} = 1

\text {Let} \quad v = e^x \Rightarrow \frac{dv}{dx} = e^x

\Rightarrow I = \left[ x e^x - \int e^x \, dx \right]_1^2 = \left[ x e^x - e^x \right]_1^2

= \left[ 2(e^2 - e^2) \right] - \left(\cancel {e - e} \quad\right)

= e^2

Problem Statement:

Note

By first using the substitution $t = \sqrt{x+1}$ , find $\int e^{2\sqrt{x+1}} \, dx$ .

Solution:

\text {Let} \quad t = (x+1)^{\frac{1}{2}}

\Rightarrow \frac{dt}{dx} = \frac{1}{2} (x+1)^{-\frac{1}{2}} = \frac {1}{2(x+1)^{\frac{1}{2}}} = \frac{1}{2t}

\Rightarrow \frac{dt}{dx} = \frac{1}{2t}

\Rightarrow dt = \frac{1}{2t} dx \Rightarrow dx = 2t \, dt

Substitution:

I = \int e^{2t} \times 2t \, dt = \int 2te^{2t} \, dt

Integrating by parts:

\text {Let} \quad u = 2t \Rightarrow \frac{du}{dt} = 2

\text {Let} \quad v = \frac{1}{2}e^{2t} \Rightarrow \frac{dv}{dt} = e^{2t}

I = \cancel2t \times \frac {1}{\cancel2}e^{2t}-\int \frac {1}{\cancel2}e^{2t} \times \cancel2 \, dt

I = t e^{2t} - \int e^{2t} \, dt

= t e^{2t} - \frac{1}{2} e^{2t} + c

\boxed {= \sqrt{x+1} e^{2\sqrt{x+1}} - \frac{1}{2} e^{2\sqrt{x+1}} + c}

:::

Integration by Parts (AQA A-Level Mathematics): Revision Notes

8.2.8 Integration by Parts