Integration of sec(v+π/2)tan(v+π/2) by substitution

[fstam2]

I am going crazy on this problem:

$$\int sec(v+(\pi/2)) tan(v+\pi/2)) dv$$

if I substitute u= $$tan(v+\pi/2)) dv$$, can I use the product rule to find du= $$sec(v+(\pi/2)) dv$$.

Thanks, Todd

 

[Galileo]

Use $\sin(x+\pi/2)=\cos(x)$ and $\cos(x+\pi/2)=-\sin(x)$ to rewrite the integrand. Then subsitute $u=\frac{1}{\sin(x)}$.

 

[dextercioby]

Else,use the definition and the substitution $$x+\frac{\pi}{2}=u$$...It's really simple.

And another one:
$$d[\sec(x+\frac{\pi}{2})]=\sec(x+\frac{\pi}{2})\tan(x+\frac{\pi}{2})dx$$

so the integration is immediate...

Daniel.