Question

1/(secx-tanx)^2 integral

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ \left( \sec x -\tan x \right)^{2} } dx\]

Answer 2

\[\frac{ 1 }{ (\sec x - \tan x)^2 }=\frac{ 1 }{ (\frac{ 1 }{ \cos x } - \frac{ \sin x }{ \cos x })^2 }=\]
\[\frac{ \cos ^2 x }{ (1-\sin x )^2 }\]
\[=\frac{ \cos ^2 x }{ (1-\sin x )^2 } \times \frac{ (1+\sin x )^2 }{ (1+\sin x )^2 }\]
\[=\frac{ \cos ^2 x (1+\sin x)^2 }{ (1-\sin^2x)^2) }=\]
\[=\frac{ \cos ^2 x (1+\sin x)^2 }{ (\cos^2x)^2 }=\frac{ (1+\sin x)^2 }{ \cos ^2x }=\]
\[\frac{ (1+\sin x)^2 }{ \cos ^2x }=\frac{ (1+\sin^2 x +2 \sin x) }{ \cos ^2x }=\]
\[\frac{ 1 }{ \cos ^2x }+\frac{ \sin ^2 }{ \cos ^2x }+\frac{ 2 \sin x }{ \cos ^2x }=\]
now it is easy to integrate

Answer 3

\[\int\limits_{?}^{?}(\frac{ 1 }{\cos ^2x }+\frac{ \sin ^2 x }{\cos ^2x }+\frac{ 2isnx }{ \cos ^2 x })dx\]

Answer 4

\[\int\limits (\tan ^2x+1)dx +\int\limits (\tan ^2x)dx+2\int\limits (\frac{ sinx }{\cos ^2 x })dx\]

Answer 5

1/cos^2 x + sin^2 x/cos^2 x + 2 sinx/cos^2 x
= sec^2 x + tan^2 x + 2sinx/cos * 1/cosx
= sec^2 x + (sec^2 - 1) + 2secxtanx
= 2 sec^2 x + 2 secxtanx - 1

see that int sec^2 x = tan(x)
and
int secxtanx = secx

Answer 6

I just thought of another way to do this one (without having to convert everything to sine/cosine). Note that \(\tan^2x+1=\sec^2 x\implies \color{red}{\sec^2x - \tan^2 x}=\color{red}{1}\). Therefore,
\[\large\begin{aligned} \int\frac{\color{red}{1}}{(\sec x-\tan x)^2}\,dx &= \int\frac{\color{red}{\sec^2x -\tan^2 x}}{(\sec x-\tan x)^2}\,dx \\ &= \int\frac{(\sec x+\tan x)(\sec x-\tan x)}{(\sec x-\tan x)^2}\,dx \\ &= \int\frac{\sec x+\tan x}{\sec x-\tan x}\,dx \\ &= \int \frac{(\sec x+\tan x)^2}{(\sec x-\tan x)(\sec x+\tan x)}\,dx\\ &=\int\frac{\sec^2x + 2\sec x\tan x+\tan^2 x}{\color{red}{\sec^2x-\tan^2 x}}\,dx \\ &= \int \sec^2 x+2\sec x\tan x+\tan^2 x\,dx\\ &= \int 2\sec^2 x + 2\sec x\tan x - 1\,dx\\ &= 2\tan x + 2\sec x-x + C \end{aligned}\]which matches the answer you'll get using the other method that was presented. :-)

Answer 7

omg thank you so much

Answer 8

After doing it the way I did, I realized it can be done in even fewer steps. XD

Still keeping in mind that \(\tan^2 x+1 = \sec^2x\implies \sec^2 x - \tan^2 x = 1\), note that
\[\large \begin{aligned}\int \frac{1}{(\sec x-\tan x)^2}\,dx &= \int\frac{(\sec x+\tan x)^2}{(\sec x-\tan x)^2(\sec x+\tan x)^2}\,dx\\ &= \int\frac{\sec^2 x+2\sec x\tan x + \tan^2x}{[(\sec x-\tan x)(\sec x+\tan x)]^2}\,dx \\ &= \int\frac{2\sec^2 x+2\sec x\tan x -1}{(\sec^2x-\tan^2 x)^2}\,dx \\ &= \int 2\sec^2 x+2\sec x\tan x -1\,dx\\ &= 2\tan x + 2\sec x - x + C\end{aligned}\]