Answer to Question #167963 in Calculus for lauren

Question

a)Find the antiderivative for sec(x)tan(x)dx

b)Given: integral from 0 to pi sec(x)+tan(x)dx-->If you found an antiderivative in part (a), explain why the result for the definite integral using the same function is undefined.

Answer 1

a) $\int \sec x \tan x \,dx = \int \dfrac{1}{\cos x}\cdot \dfrac{\sin x}{\cos x}\,dx = \int \dfrac{\sin x}{\cos^2 x}\, dx = \Big | y = \cos x , dy = -\sin x\,dx \Big| = - \int \dfrac{1}{y^2}\,dy = \dfrac{1}{y} + C = \dfrac{1}{\cos x} = C.$

Here we see the arbitrary constant. If we want to obtain the definite integral, we may use any constant C because it reduces. But there may be a problem if we want to obtain the definite integral with upper (or lower) limit equal to $\pi/2,$ because $\cos (\pi/2)$ = 0.

b) $\int\limits_0^{\pi} (\sec(x)+\tan(x))\,dx = \int \limits_0^{\pi} \left(\dfrac{1}{\cos x} +\dfrac{\sin x}{\cos x} \right)\,dx = \int\limits_0^{\pi} \dfrac{1}{\cos x}\, dx+ \int\limits_{-1}^{1} \dfrac{d(\cos x)}{\cos x} = \Big( \ln \left|\tan\left(\dfrac{\pi}{4} + \dfrac{x}{2}\right)\right| - \dfrac{1}{\cos x} \Big) \Big |_0^{\pi}.$

The principal value of the integral is 0.