Question

Find the integral of
∫secu du

Answer 1

Answer on Question #50830 - Math - Integral Calculus

Question. Find indefinite integral

$$\int \sec u \, du$$

Solution. Recall that by definition $\sec u = \frac{1}{\cos u}$. Thus

$$\int \sec u \, du = \int \frac{du}{\cos u}.$$

Let us make the following change of variables. Denote

$$t = \tan(u/2).$$

Then

$$\begin{aligned} \cos u &= \cos 2(u/2) = \cos^2(u/2) - \sin^2(u/2) \\ &\quad \left( \text{use the relation } \cos^2(u/2) + \sin^2(u/2) = 1 \right) \\ &= \frac{\cos^2(u/2) - \sin^2(u/2)}{\cos^2(u/2) + \sin^2(u/2)} \\ &\quad \left( \text{divide numerator and denominator by } \cos^2(u/2) \right) \\ &= \frac{1 - \tan^2(u/2)}{1 + \tan^2(u/2)} = \frac{1 - t^2}{1 + t^2}. \end{aligned}$$

Moreover, $u = 2\arctan t$, whence

$$du = (2\arctan t)'\,dt = \frac{2dt}{1 + t^2}.$$

Substituting these in the integral we get

$$\begin{aligned} \int \sec u \, du &= \int \frac{du}{\cos u} = \int \frac{2dt}{1 + t^2} \cdot \frac{1 + t^2}{1 - t^2} \\ &= \int \frac{2dt}{1 - t^2} = \int \frac{2dt}{(1 - t)(1 + t)} = \int \frac{dt}{1 - t} + \int \frac{dt}{1 + t} \\ &= \ln |1 - t| + \ln |1 + t| + C = \ln |(1 - t)(1 + t)| + C = \ln |(1 - t^2)| + C \\ &= \ln |(1 - \tan^2(u/2)| + C). \end{aligned}$$

Answer. $\int \sec u \, du = \ln |(1 - \tan^2(u/2)| + C.$

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