Further Pure 1 - Further Calculus - Weierstrass Substitution

We know from the further trigonometry section that if we set:

\displaystyle t = \tan \frac x 2

(for x \ne \pi, 3 \pi, \ldots) then we have:

\displaystyle \sin x = \frac {2t} {1 + t^2}

and:

\displaystyle \cos x = \frac {1 - t^2} {1 + t^2}

We can use therefore use the substitution \displaystyle t = \tan \frac x 2 to rewrite trigonometric integrals as integrals of rational functions, (ie. fractions involving polynomials) which can succumb to straightforward integration techniques, like partial fractions.

Note that we have:

\displaystyle \frac {\mathrm dt} {\mathrm dx} = \frac 1 2 \sec^2 \frac x 2

To write this in terms of t, remember the identity:

\displaystyle 1 + \tan^2 \theta = \sec^2 \theta

Setting \theta = \dfrac x 2 and dividing by 2 we have:

\displaystyle \frac 1 2 \sec^2 \frac x 2 = \frac 1 2 \left(1 + \tan^2 \frac x 2\right) = \frac 1 2 \left(1 + t^2\right)

So we have:

\displaystyle \frac {\mathrm dt} {\mathrm dx} = \frac 1 2 \left(1 + t^2\right)

and so, informally:

\displaystyle \mathrm dx = \frac 2 {1 + t^2} \mathrm dt

We therefore have that, using integration by substitution:

\displaystyle \int f(\cos \theta, \sin \theta) \mathrm d\theta = 2 \int f \left(\frac {1 - t^2} {1 + t^2}, \frac {2t} {1 + t^2}\right) \frac {\mathrm dt} {1 + t^2}

for any suitable function f.

To take an explicit example, consider for example:

\displaystyle \int \frac 1 {\sin \theta + \cos \theta} \mathrm d\theta

Substituting \displaystyle t = \tan \frac x 2 we get:

\begin{align*}\int \frac 1 {\sin \theta + \cos \theta} \mathrm d\theta & = \int \frac 1 {\frac {2t} {1 + t^2} + \frac {1 - t^2} {1 + t^2}} \times \frac 2 {1 + t^2} \mathrm dt \\ & = 2 \int \frac 1 {2t + 1 - t^2} \mathrm dt\end{align*}

To evaluate this new integral, we can complete the square and write:

\begin{align*}2t + 1 - t^2 & = -(t^2 - 2t - 1) \\ & = -\left((t - 1)^2 - 2\right) \\ & = 2 - (t - 1)^2\end{align*}

So we have:

\displaystyle 2 \int \frac 1 {2 - (t - 1)^2} \mathrm dt

We want to apply the identity:

\displaystyle 2 \int \frac 1 {a^2 - x^2} \mathrm dx = \frac 1 a \ln \left|\frac {a + x} {a - x}\right| + C

Setting u = t - 1 our integral becomes:

\displaystyle 2 \int \frac 1 {\left(\sqrt 2\right)^2 - u^2} \mathrm du

So applying the above identity we have:

\begin{align*}\int \frac 1 {\sin \theta + \cos \theta} \mathrm d\theta & = 2 \int \frac 1 {\left(\sqrt 2\right)^2 - u^2} \mathrm du \\ & = \frac 1 {\sqrt 2} \ln \left|\frac {\sqrt 2 + u} {\sqrt 2 - u}\right| + C \\ & = \frac 1 {\sqrt 2} \ln \left|\frac {\sqrt 2 + t - 1} {\sqrt 2 - t + 1}\right| + C \\ & = \boxed{\frac 1 {\sqrt 2} \ln \left|\frac {\sqrt 2 + \tan \frac x 2 - 1} {\sqrt 2 - \tan \frac x 2 + 1}\right| + C}\end{align*}

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