How to find a primitivefunction for $(\sec(x))^3$ I had a test where I was asked this question. I integrated the function by taking $\tan x=u$ and then $\sec^2x \, dx=du$. I took $\sec(x) =\sqrt{1+u^2}$ and then using the basic formulas calculated the correct answer but my teacher didn't give me marks for my solution. Please  explain me why. He gave me a zero.I can't post pictures otherwise I would have shown you 
I got the answer $(\tan x\sec x-\log(\tan x+\sec x)/2$
Here the picture of my solution https://drive.google.com/file/d/1E9e1JoX_8C28h7k8uW6bBpoYZFy1LuCM/view?usp=drivesdk
 A: \begin{align}
\int\sec^3x\,dx 
&= \int\dfrac{1}{\cos^3x}\,dx \\
&= \int\dfrac{\cos x}{(1-\sin^2x)^2}\,dx \hspace{0.5cm};\hspace{0.5cm} \sin x=u\\
&= \int\dfrac{1}{(1-u^2)^2}\,dx \\
&= \dfrac{u}{2(1-u^2)}+\dfrac{1}{4}\ln\dfrac{1+u}{1-u}+C\\
&= \dfrac{u}{2(1-u^2)}+\dfrac{1}{4}\ln\dfrac{1+u}{1-u}+C\\
&= \dfrac{\sin x}{2\cos^2x}+\dfrac{1}{4}\ln\dfrac{1+\sin x}{1-\sin x}+C
\end{align}
A: Here's one way:
\begin{align}
\int \sec^3 x \, dx = {} & \int (\sec x) \Big( \sec^2 x \, dx\Big) = \overbrace{\int u\, dv = uv - \int v \, du}^{\text{integration by parts}} \\[10pt]
= {} & \sec x \tan x - \int (\tan x) \big( \sec x\tan x\, dx\big) \\[10pt]
= {} & \sec x \tan x - \int(\sec x)(\tan^2 x) \, dx \\[10pt]
= {} & \sec x \tan x - \int(\sec x) (\sec^2 x - 1) \,dx \\[10pt]
= {} & \sec x\tan x + \int \sec x\,dx - \int\sec^3 x\,dx. \\[10pt]
& \text{Then adding $\int\sec^3 x\,dx$ to both sides, we get} \\[10pt]
2\int\sec^3 x\, dx & = \sec x \tan x + \int \sec x\, dx. \\[10pt]
& \text{Hence} \\[10pt]
\int \sec^3 x\, dx & = \frac 1 2 \sec x \tan x + \frac 1 2 \int \sec x\, dx = \cdots
\end{align}
