mikewill54
New member
- Joined
- Mar 7, 2016
- Messages
- 31
Hi,
Can someone help with this question, im using the guess and check method for integration from the calculus for dummies workbook
But im not getting their answer when I differentiate [math]sec(5t-pi)[/math]
[math]sec(5T-pi)\\ \\Re-Write \ :sec(t)= \frac{1}{cosT} \\ =\frac{1}{cos(5T-pi)} \\ \underline{Use \ angle \ difference \ identity} \\Cos(s-t)=cos(s)cos(t)+sin(s)sin(t) \\=\frac{1}{Cos(5t)(Cos(pi)+Sin(5t)Sin(pi)} \\Cos(pi)=-1 \\Sin(pi)=0 \\=-\frac{1}{Cos(5t)} \\ \underline{Trig \ identity} \\ \frac{1}{cos(x)}=sec(x) \\So \\=\frac{d}{dx}(-sec(5x) \\-\frac{d}{dx}(sec(5x)) \\Use \ the \ chain \ rule \\f =sec(u) \ \ u=5x \\\frac{d}{du}sec(u) = sec(u)tan(u) \\\frac{d}{dx}=5 \\Substitute \ back \ in \\5.Sec(5x)tan(5x)[/math]
Thanks for any help
Regards
Mike
Can someone help with this question, im using the guess and check method for integration from the calculus for dummies workbook
But im not getting their answer when I differentiate [math]sec(5t-pi)[/math]
[math]sec(5T-pi)\\ \\Re-Write \ :sec(t)= \frac{1}{cosT} \\ =\frac{1}{cos(5T-pi)} \\ \underline{Use \ angle \ difference \ identity} \\Cos(s-t)=cos(s)cos(t)+sin(s)sin(t) \\=\frac{1}{Cos(5t)(Cos(pi)+Sin(5t)Sin(pi)} \\Cos(pi)=-1 \\Sin(pi)=0 \\=-\frac{1}{Cos(5t)} \\ \underline{Trig \ identity} \\ \frac{1}{cos(x)}=sec(x) \\So \\=\frac{d}{dx}(-sec(5x) \\-\frac{d}{dx}(sec(5x)) \\Use \ the \ chain \ rule \\f =sec(u) \ \ u=5x \\\frac{d}{du}sec(u) = sec(u)tan(u) \\\frac{d}{dx}=5 \\Substitute \ back \ in \\5.Sec(5x)tan(5x)[/math]
Thanks for any help
Regards
Mike