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Integration of sec x

Integration of sec x can be found using various integration technique like integration by substitution, Integration by partial fraction along with trigonometric identities. The various formula for integration of sec x are

I

sec(x)dx=ln|sec(x)+tan(x)|+C

II

\int \sec(x) \, dx = -\ln | \sec(x) – \tan(x) | + C

III

\int \csc(x) \, dx = \ln | \tan ( \frac {x}{2} + \frac {\pi}{2})| + C

IV

\int \csc(x) \, dx = \frac {1}{2} \ln | \frac {1 + \sin x}{1 – \sin x} | + C

Lets check out the proof of each of these

Proof of I

Integral of secant can be little tricky and can be solved by multiplying and dividing by (\sec(x) + \tan(x)). Here’s how it’s done:

\int \sec(x) \, dx = \int \sec(x) \cdot \frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} \, dx \\ = \int \frac{\sec^2(x) + \sec(x)\tan(x)}{\sec(x) + \tan(x)} \, dx.

Now, taking u as
u= \sec(x) + \tan(x)
du =(\sec(x)\tan(x) + \sec^2(x)) dx

Therefore integration becomes

\int \sec(x) \, dx =\int \frac {1}{u} \, du,

hence

\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C

Proof of II

We can write the result of I as

\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C = \ln | \frac {\sec^2(x) – \tan^2(x)}{\sin(x) – \sin(x) } | + C \\ =\ln | \frac {1}{\sec(x) – \tan(x) } | + C = \ln | (\sec(x) – \tan(x))^{-1} | + C \\ =- \ln | \sec(x) – \tan(x) | + C

Proof of III

This can be provided using trigonometric formulas

\int \sec(x) \, dx =\int \frac {1}{\cos(x)} \, dx =\int \frac {1}{\sin(x + \frac {\pi}{2})} \, dx =\int \frac {1}{2 \sin(x/2 + \pi/4) \cos(x/2 + \pi/4)} \, dx

Dividing and Multiplying cos(x/2 + \pi/4) in the denominator

\int \sec(x) \, dx = \frac {1}{2}\int \frac {\sec^2(x/2+ \pi/4)}{\tan(x/2+ \pi/4)} \, dx

Lets u=\tan(x/2 + \pi/4)
du= \frac{1}{2} \sec^2(x/2 + \pi/4) dx

Therefore

\int \sec(x) \, dx = \int \frac {1}{u} \, du = ln |u| + C

Hence

\int \sec(x) \, dx = \ln | \tan (\frac {x}{2} +\frac {\pi}{2}) | + C

Proof of IV

This can be proof by using integration by partial fractions

\int \sec(x) \, dx =\int \frac {1}{\cos(x)} \, dx =\int \frac {cos(x)}{\cos^2(x)} \, dx =\int \frac {cos(x)}{ 1- \sin^2(x)} \, dx

Lets u=\sin(x)
du= \cos(x) dx

Therefore

\int \sec(x) \, dx =\int \frac {1}{1 -u^2} \, du =\frac {1}{2} \int [\frac {1}{1+u} + \frac {1}{1-u} ]\, du =\frac {1}{2} ln |\frac {1+u}{1-u}| + C

Hence

\int \csc(x) \, dx = \frac {1}{2} \ln | \frac {1 + \sin x}{ 1- \sin x} | + C

Solved Example

Example 1

\int_{0}^{\pi/4} \sec(x) \, dx.

Solution

Using the result from above:

\int_{0}^{\pi/4} \sec(x) \, dx = \left[ \ln|\sec(x) + \tan(x)| \right]_{0}^{\pi/4}.

Evaluating at the bounds:

\begin{align} \left[ \ln|\sec(x) + \tan(x)| \right]_{0}^{\pi/4} &= \ln|\sec(\pi/4) + \tan(\pi/4)| – \ln|\sec(0) + \tan(0)| \\ &= \ln|(\sqrt{2}) + 1| – \ln|1 + 0| \\ &= \ln(\sqrt{2} + 1). \end{align}

So, \int_{0}^{\pi/4} \sec(x) \, dx = \ln(\sqrt{2} + 1).

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