maths > integral-calculus

Integration using Identities

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Overview

Integration using Identities :

when $f\left(x\right)={\Pi }_{i=1}^{n}{g}_{i}\left(x\right)$$f \left(x\right) = {\Pi}_{i = 1}^{n} {g}_{i} \left(x\right)$, use identities to convert to $f\left(x\right)=\sum _{i=1}^{m}{h}_{i}\left(x\right)$$f \left(x\right) = {\sum}_{i = 1}^{m} {h}_{i} \left(x\right)$, such that integration can be individually performed on ${h}_{i}\left(x\right)$${h}_{i} \left(x\right)$

identity

Consider the integration $\int \left(x+2\right)\left(x+3\right)dx$$\int \left(x + 2\right) \left(x + 3\right) \mathrm{dx}$. To work out the integral use an algebraic identity to convert the multiplication to a sum of terms.

$\int \left(x+2\right)\left(x+3\right)dx$$\int \left(x + 2\right) \left(x + 3\right) \mathrm{dx}$.

use the identity $\left(x+a\right)\left(x+b\right)$$\left(x + a\right) \left(x + b\right)$$={x}^{2}+\left(a+b\right)x+ab$$= {x}^{2} + \left(a + b\right) x + a b$
$=\int \left[{x}^{2}+\left(2+3\right)x+2×3\right]dx$$= \int \left[{x}^{2} + \left(2 + 3\right) x + 2 \times 3\right] \mathrm{dx}$

This integral can be computed for each of the terms.

cos squared

Consider the integration $\int {\mathrm{cos}}^{2}xdx$$\int {\cos}^{2} x \mathrm{dx}$.

We know the trigonometric identity $\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$

substitute ${\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x$${\sin}^{2} x = 1 - {\cos}^{2} x$ and rearrange the terms
${\mathrm{cos}}^{2}x=\frac{1+\mathrm{cos}2x}{2}$${\cos}^{2} x = \frac{1 + \cos 2 x}{2}$

$\int {\mathrm{cos}}^{2}xdx$$\int {\cos}^{2} x \mathrm{dx}$

$=\int \left[\frac{1}{2}+\frac{\mathrm{cos}2x}{2}\right]dx$$= \int \left[\frac{1}{2} + \frac{\cos 2 x}{2}\right] \mathrm{dx}$

This integral can be computed for each of the terms.

some identities

Some trigonometric identities useful for integration are

$2\mathrm{sin}x\mathrm{cos}y$$2 \sin x \cos y$$=\mathrm{sin}\left(x+y\right)+\mathrm{sin}\left(x-y\right)$$= \sin \left(x + y\right) + \sin \left(x - y\right)$
$2\mathrm{sin}x\mathrm{cos}x$$2 \sin x \cos x$$=\mathrm{sin}\left(x+y\right)$$= \sin \left(x + y\right)$

$2\mathrm{cos}x\mathrm{cos}y$$2 \cos x \cos y$$=\mathrm{cos}\left(x-y\right)+\mathrm{cos}\left(x+y\right)$$= \cos \left(x - y\right) + \cos \left(x + y\right)$
$2{\mathrm{cos}}^{2}x$$2 {\cos}^{2} x$$=1+\mathrm{cos}\left(2x\right)$$= 1 + \cos \left(2 x\right)$
$4{\mathrm{cos}}^{3}x$$4 {\cos}^{3} x$$=3\mathrm{cos}x+\mathrm{cos}3x$$= 3 \cos x + \cos 3 x$

$2\mathrm{sin}x\mathrm{sin}y$$2 \sin x \sin y$$=\mathrm{cos}\left(x-y\right)-\mathrm{cos}\left(x+y\right)$$= \cos \left(x - y\right) - \cos \left(x + y\right)$
$2{\mathrm{sin}}^{2}x$$2 {\sin}^{2} x$$=1-\mathrm{cos}\left(2x\right)$$= 1 - \cos \left(2 x\right)$
${\mathrm{sin}}^{3}x$${\sin}^{3} x$$=3\mathrm{sin}x-\mathrm{sin}3x$$= 3 \sin x - \sin 3 x$

example

Integrate $\int {\mathrm{cos}}^{2}xdx+\int {\mathrm{sin}}^{2}xdx$$\int {\cos}^{2} x \mathrm{dx} + \int {\sin}^{2} x \mathrm{dx}$.

This can be integrated using the identity ${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1$${\cos}^{2} x + {\sin}^{2} x = 1$

summary

Integration using Identities: when $f\left(x\right)={\Pi }_{i=1}^{n}{g}_{i}\left(x\right)$$f \left(x\right) = {\Pi}_{i = 1}^{n} {g}_{i} \left(x\right)$ Use identities to convert to $f\left(x\right)=\sum _{i=1}^{m}{h}_{i}\left(x\right)$$f \left(x\right) = {\sum}_{i = 1}^{m} {h}_{i} \left(x\right)$, such that integration can be individually performed on ${h}_{i}\left(x\right)$${h}_{i} \left(x\right)$.

Outline