# How do you integrate int sec^2(x/2)tan(x/2)?

Dec 8, 2016

${\tan}^{2} \left(\frac{x}{2}\right) + {C}_{1} \text{ }$or ${\sec}^{2} \left(\frac{x}{2}\right) + {C}_{2}$

#### Explanation:

regonising the function and its derivative is important.

method 1

$\frac{d}{\mathrm{dx}} {\tan}^{2} \left(\frac{x}{2}\right) = 2 \times \left(\frac{1}{2}\right) \tan \left(\frac{x}{2}\right) {\sec}^{2} \left(\frac{x}{2}\right)$

so$\text{ } \int {\sec}^{2} \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right) \mathrm{dx} = {\tan}^{2} \left(\frac{x}{2}\right) + {C}_{1}$

method 2

$\frac{d}{\mathrm{dx}} {\sec}^{2} \left(\frac{x}{2}\right) = \left(\frac{1}{2}\right) \times 2 \sec \left(\frac{x}{2}\right) \times \sec \left(\frac{x}{2}\right) \times \tan \left(\frac{x}{2}\right)$

$= {\sec}^{2} \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$

so$\text{ } \int {\sec}^{2} \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right) \mathrm{dx} = {\sec}^{2} \left(\frac{x}{2}\right) + {C}_{2}$

the two solutions can be shown to be consistent by
using the identity

$1 + {\tan}^{2} \theta + {\sec}^{2} \theta$

Dec 8, 2016

Method 1
If we recognize that $\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$, then we might try the substitution

$u = \tan \left(\frac{x}{2}\right)$.

This makes $\mathrm{du} = \frac{1}{2} {\sec}^{2} \left(\frac{x}{2}\right) \mathrm{dx}$, and the integral becomes

$2 \int u \mathrm{du} = {u}^{2} + C = {\tan}^{2} \left(\frac{x}{2}\right) + C$

Method 2

If we recognize that $\frac{d}{\mathrm{dx}} \left(\sec x\right) = \sec x \tan x$, then we might try the substitution

$u = \sec \left(\frac{x}{2}\right)$.

This makes $\mathrm{du} = \frac{1}{2} \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right) \mathrm{dx}$, and the integral becomes

$2 \int u \mathrm{du} = {u}^{2} + C = {\sec}^{2} \left(\frac{x}{2}\right) + C$

Final Note

A student who has not seem such a result may wonder, "How can both answers be right?"

The answer is in the constant, $C$.

Recall that ${\tan}^{2} x + 1 = {\sec}^{2} x$

So if we have
${\tan}^{2} \left(\frac{x}{2}\right) + 5$ we can rewrite this as ${\sec}^{2} \left(\frac{x}{2}\right) + 4$.

Our "two" answers have different $C$'s.