Question

∫√(secx-1)dx

Answer 1

What are your fail attempts?

Answer 2

\[\begin{align*}\int\sqrt{\sec x-1}~dx&=\int\sqrt{\frac{1}{\cos x}-1}~dx\\
&=\int\sqrt{\frac{1-\cos x}{\cos x}}~dx\\
&=\int\sqrt{\frac{1-\cos x}{\cos x}}\cdot\sqrt{\frac{1+\cos x}{1+\cos x}}~dx\\
&=\int\sqrt{\frac{1-\cos^2 x}{\cos x(1+\cos x)}}~dx\\
&=\int\sqrt{\frac{\sin^2 x}{\cos x(1+\cos x)}}~dx\\
&=\int\frac{\sin x}{\sqrt{\cos x(1+\cos x)}}~dx
\end{align*}\]
Let \(u=\cos x\), then \(-du=\sin x~dx\), so you have
\[\begin{align*}\cdots&=-\int\frac{du}{\sqrt{u(1+u)}}\\
&=-\int\frac{du}{\sqrt{u^2+u}}
\end{align*}\]
Complete the square in the denominator:
\[u^2+u=u^2+u+\frac{1}{4}-\frac{1}{4}=\left(u+\frac{1}{2}\right)^2-\frac{1}{4}\]
Now substitute \(t=u+\dfrac{1}{2}\), so \(dt=du\):
\[\begin{align*}\cdots&=-\int\frac{dt}{\sqrt{t^2-\dfrac{1}{4}}}
\end{align*}\]
One last substituon... Let \(t=\dfrac{1}{2}\sec s\), so that \(dt=\dfrac{1}{2}\sec s\tan s~ds\).
\[\begin{align*}\cdots&=-\int\frac{\dfrac{1}{2}\sec s\tan s}{\sqrt{\left(\dfrac{1}{2}\sec s\right)^2-\dfrac{1}{4}}}~ds\\
&=-\int\frac{\dfrac{1}{2}\sec s\tan s}{\sqrt{\dfrac{1}{4}\sec^2 s-\dfrac{1}{4}}}~ds\\
&=-\int\frac{\sec s\tan s}{\sqrt{\sec^2 s-1}}~ds\\
&=-\int\frac{\sec s\tan s}{\sqrt{\tan^2s}}~ds\\
&=-\int\sec s~ds
\end{align*}\]

Answer 3

Hmm, this doesn't get the right answer though... I guess you can't just say \(\sqrt{\sin^2x}=\sin x\)...

Answer 4

Well, at any rate, you have \(\sqrt{\sin^2x}=|\sin x|\), so you might just have to split this up into cases for when \(\sin x<0\) and \(\sin x>0\).