What an integral

The integral is the famous integral of secant cubed .

The integral can be done by parts, using this formula. $\int u\text{ }dv=uv-\int v\text{ }du$

Let $u=\sec x,$ $du=\sec x\tan x\text{ }dx,$ $v=\tan x,$ and $dv=\sec^2x\text{ }dx.$ Plugging in to the formula, $\begin{aligned} \int\sec^3x\text{ }dx&=\sec x\tan x-\int \sec x\tan^2x\text{ }dx\\ &=\sec x\tan x-\int\sec x(\sec^2x-1)\text{ }dx\\ &=\sec x\tan x-\int\sec^3x\text{ }dx+\int\sec x\text{ }dx \end{aligned}$ Move the integral of $\sec^3x$ back over to the LHS. $\begin{aligned} 2\int\sec^3x\text{ }dx&=\sec x\tan x+\int\sec x\text{ }dx\\ &=\sec x\tan x+\ln|\sec x+\tan x|+C \end{aligned}$ Divide each side by $2$ to get the final answer. $\int\sec^3x\text{ }dx=\dfrac{1}{2}(\sec x\tan x+\ln|\sec x+\tan x|)+C$

Rearranging of the trigonometric functions shows this. $\dfrac{1}{2}(\sec x\tan x+\ln|\sec x+\tan x|)+C=\dfrac{1}{2}\left(\dfrac{\sin x}{\cos^2x}+\ln\left(\dfrac{1+\sin x}{\cos x}\right)\right)+C$ Thus, $\gamma=1,$ $\psi=2,$ and $\alpha=1,$ so $\gamma+\psi+\alpha-1=\boxed{3}$

Sorry for the incorrect answer @Trevor B. .. Thats embarrassing careless me, how did you know the incorrect answer was 3 any ways This problem can also be solved with out integration by parts let $u=cosx$ , and $s=sinx$ $=\int { \frac { 1 }{ { u }^{ 3 } } } \\ =\int { \frac { { u } }{ { u }^{ 4 } } } \\ =\int { \frac { u }{ { \left( 1-{ s }^{ 2 } \right) }^{ 2 } } } \\ =\int { \frac { 1 }{ 4 } u{ \left( \frac { 1 }{ 1+s } +\frac { 1 }{ 1-s } \right) }^{ 2 } } \\ =\frac { 1 }{ 4 } \int { \left( \frac { u }{ { \left( 1+s \right) }^{ 2 } } +\frac { u }{ { \left( 1-s \right) }^{ 2 } } \right) } +\frac { 1 }{ 2 } \int { \frac { u }{ { \left( 1-s \right) }^{ 2 } } } \\ =\frac { 1 }{ 4 } \left( \frac { -1 }{ 1+s } +\frac { 1 }{ 1-s } \right) +\frac { 1 }{ 4 } \int { \left( \frac { u }{ 1+s } +\frac { u }{ 1-s } \right) } \\ =\frac { s }{ 2{ u }^{ 2 } } +\frac { 1 }{ 4 } ln\left( \frac { 1+s }{ 1-s } \right) \\ =\frac { 1 }{ 2 } \left( \frac { s }{ { u }^{ 2 } } +ln\left( \frac { 1+s }{ u } \right) \right) \\ =\frac { 1 }{ 2 } \left( \frac { sinx }{ { cos }^{ 2 }x } +ln\left( \frac { 1+sinx }{ cosx } \right) \right)$