MHB Integral of sec(x): Solving with Substitution

[Mathick]

Hello! I need to compute the following integral: $ \int \sec\left({x}\right) \; dx $ using the substitution $ t=\tan\left({\frac{x}{2}}\right) $ what I did. HOWEVER, they ask to compute the same integral knowing that $ \int \csc\left({x}\right) \; dx = \log_{e}\left({\tan\left({\frac{x}{2}}\right)}\right) + C $ and using substitution $ y=\frac{\pi}{2} - x $. And I have a lot of problems with the second part of this task. I would be very grateful for some help or advice.

 

[Greg]

Use the identity $\cos\left(\frac{\pi}{2}-z\right)=\sin(z)$.

 

[soroban]

I need to compute the following integral: $\displaystyle \int \sec\left({x}\right) \; dx $ using the substitution $ z=\tan\left({\frac{x}{2}}\right) $

z = \tan\frac{x}{2} \quad\Rightarrow\quad x = 2\arctan z \quad\Rightarrow\quad dx = \frac{2\,dz}{1+z^2} \quad\Rightarrow\quad \cos x = \frac{1-z^2}{1+z^2}

\displaystyle \int\sec x\,dx \;=\;\int \frac{1+z^2}{1-z^2}\,\frac{2\,dz}{1+z^2} \;=\; \int\frac{2\,dz}{1-z^2} \;=\;\int\left(\frac{1}{1-z} + \frac{1}{1+z}\right)dz

. . . =\;-\ln(1-z) + \ln(1+z) +C \;=\;\ln\left|\frac{1+z}{1-z}\right| + C \;=\; \ln\left|\frac{1+\tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\right| + C

. . . =\;\ln\left|\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right| + C \;=\;\frac{1}{2}\ln\left|\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right|^2 + C

. . . =\; \frac{1}{2}\ln\left|\frac{\sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} + \cos^2\frac{2}{2}} {\cos^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} + \sin^2\frac{x}{2}}\right| + C \;=\;\frac{1}{2}\ln \left|<br /> \frac{1 + \sin x}{1-\sin x}\right| + C

. . . =\; \frac{1}{2}\ln\left|\frac{1+\sin x}{1 - \sin x}\cdot\frac{1 + \sin x}{1 + \sin x}\right| + C \;=\;\frac{1}{2}\ln\left|\frac{(1+\sin x)^2}{1-\sin^2x}\right| + C<br />

. . . =\; \frac{1}{2}\ln\left|\frac{(1+\sin x)^2}{\cos^2x}\right| + C \;=\;\ln\left|\frac{(1+\sin x)^2}{\cos^2x}\right|^{\frac{1}{2}} + C

. . . =\;\ln\left|\frac{1+\sin x}{\cos x}\right| + C \;=\; \ln\left|\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right| + C \;=\;\ln|\sec x + \tan x| + C

 

[Mathick]

z = \tan\frac{x}{2} \quad\Rightarrow\quad x = 2\arctan z \quad\Rightarrow\quad dx = \frac{2\,dz}{1+z^2} \quad\Rightarrow\quad \cos x = \frac{1-z^2}{1+z^2}

\displaystyle \int\sec x\,dx \;=\;\int \frac{1+z^2}{1-z^2}\,\frac{2\,dz}{1+z^2} \;=\; \int\frac{2\,dz}{1-z^2} \;=\;\int\left(\frac{1}{1-z} + \frac{1}{1+z}\right)dz

. . . =\;-\ln(1-z) + \ln(1+z) +C \;=\;\ln\left|\frac{1+z}{1-z}\right| + C \;=\; \ln\left|\frac{1+\tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\right| + C

. . . =\;\ln\left|\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right| + C \;=\;\frac{1}{2}\ln\left|\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right|^2 + C

. . . =\; \frac{1}{2}\ln\left|\frac{\sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} + \cos^2\frac{2}{2}} {\cos^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} + \sin^2\frac{x}{2}}\right| + C \;=\;\frac{1}{2}\ln \left|<br /> \frac{1 + \sin x}{1-\sin x}\right| + C

. . . =\; \frac{1}{2}\ln\left|\frac{1+\sin x}{1 - \sin x}\cdot\frac{1 + \sin x}{1 + \sin x}\right| + C \;=\;\frac{1}{2}\ln\left|\frac{(1+\sin x)^2}{1-\sin^2x}\right| + C<br />

. . . =\; \frac{1}{2}\ln\left|\frac{(1+\sin x)^2}{\cos^2x}\right| + C \;=\;\ln\left|\frac{(1+\sin x)^2}{\cos^2x}\right|^{\frac{1}{2}} + C

. . . =\;\ln\left|\frac{1+\sin x}{\cos x}\right| + C \;=\; \ln\left|\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right| + C \;=\;\ln|\sec x + \tan x| + C

I appreciate that but I didn't ask for that...

 

[I like Serena]

How about substituting $x=\frac\pi 2-y$ and $dx=-dy$, and using greg1313's suggestion?