Further integration

Integration

Standard integrals

$\frac{1}{2}\mathrm{e}^{2x + 1} + \frac{1}{3}\ln\lvert 3x + 2\rvert - \frac{1}{2}\cos 2x + \frac{1}{3}\tan 3x + c$

$\left[\frac{1}{2}\ln(2x + 1)\right]_0^1 = \frac{1}{2}\ln 3 - \frac{1}{2}\ln 1 = \frac{1}{2}\ln 3$

Using trigonometric identities

If a function cannot be integrated as it stands, change it with an identity.

$\int_0^{\frac{\pi}{4}} \tfrac{1}{2}(1 + \cos 2x)\,\mathrm{d}x = \left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_0^{\frac{\pi}{4}} = \frac{\pi}{8} + \frac{1}{4}$

The trapezium rule

The trapezium rule estimates a definite integral by splitting the area into strips of equal width $h$ and treating each as a trapezium.

$h = 1$, and the $y$-values are $1$, $\sqrt{2}$ and $\sqrt{5}$:

$\frac{1}{2}\left[1 + 2\sqrt{2} + \sqrt{5}\right] = 3.03$

Further methods (Paper 3)

Choosing a method

Inverse tangent and partial fractions

$\frac{1}{3}\tan^{-1}\frac{x}{3} + c$

Split a rational function first, then integrate each part.

The partial fractions are $\dfrac{2}{x + 1} + \dfrac{3}{x - 1}$, so the integral is $2\ln\lvert x + 1\rvert + 3\ln\lvert x - 1\rvert + c$.

A squared bracket gives a power, not a logarithm: $\displaystyle\int \frac{2}{(x + 1)^2}\,\mathrm{d}x = -\frac{2}{x + 1} + c$.

Integrals of the form $\dfrac{k\mathrm{f}'(x)}{\mathrm{f}(x)}$

Look for a top that is a multiple of the derivative of the bottom.

$\int \frac{x}{x^2 + 1}\,\mathrm{d}x = \frac{1}{2}\ln(x^2 + 1) + c$

$\int \frac{\sin x}{\cos x}\,\mathrm{d}x = -\ln\lvert\cos x\rvert + c = \ln\lvert\sec x\rvert + c$

Integration by parts

Choose $u$ to be the part that gets simpler when differentiated. The exception is a logarithm: make $u = \ln x$, since you cannot integrate it directly.

$u = x$ and $\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{2x}$, so $v = \tfrac{1}{2}\mathrm{e}^{2x}$:

$\frac{1}{2}x\mathrm{e}^{2x} - \int \frac{1}{2}\mathrm{e}^{2x}\,\mathrm{d}x = \frac{1}{2}x\mathrm{e}^{2x} - \frac{1}{4}\mathrm{e}^{2x} + c$

$u = \ln x$ and $\dfrac{\mathrm{d}v}{\mathrm{d}x} = 1$, so $v = x$:

$\left[x\ln x\right]_1^{\mathrm{e}} - \int_1^{\mathrm{e}} 1\,\mathrm{d}x = \mathrm{e} - (\mathrm{e} - 1) = 1$

Substitution

Replace $x$ everywhere, including $\mathrm{d}x$ and the limits.

$\mathrm{d}u = \mathrm{d}x$ and $x = u - 1$. The limits become $u = 1$ and $u = 4$:

$\int_1^4 (u - 1)u^{\frac{1}{2}}\,\mathrm{d}u = \left[\frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}}\right]_1^4 = \frac{2}{5}(31) - \frac{2}{3}(7) = \frac{116}{15}$