# Integrating (sinx + secx)/tanx dx

• Feb 24th 2008, 05:25 PM
L_U_K_E
Integrating (sinx + secx)/tanx dx
&int;(sinx + secx) / tanx dx

&int;(sinx/tanx)dx + &int;(secx/tanx)dx

&int;(cosx)dx + &int;(1/sinx)dx

sinx + ... not sure where to go from here. tried to go &int;(sinx/sinx^2)
1/2&int;(sinx/1-cos2x)
u=1-cos2x du=2sinxdx dx=du/2sinx
1/2&int;(sinx/2)(du/2sinx)
1/4&int;(1/u)du

so... sinx + 1/4(ln|1-cos2x|)

but the back of the book says the answer is sinx + ln | cscx-cotx |

any ideas??? help!!!!!!!!!!!!!!!!!!!!!!!!!!!
• Feb 24th 2008, 05:55 PM
Soroban
Hello, L_U_K_E!

Just one more step . . .

Quote:

$\int\frac{\sin x + \sec x}{\tan x}\,dx$

$\int\frac{\sin x}{\tan x}\,dx + \int\frac{\sec x}{\tan x}\,dx$

$\int\cos x\,dx + \int\frac{1}{\sin x}\,dx\quad{\color{blue}\Leftarrow\quad \frac{1}{\sin x} \:=\:\csc x}$

Got it?

• Feb 24th 2008, 05:59 PM
L_U_K_E
Quote:

Originally Posted by Soroban
Hello, L_U_K_E!

Just one more step . . .

Got it?

ya i got that but then i have no idea how to integrate cscx ?? can u help with that?
• Feb 24th 2008, 06:03 PM
Krizalid
Multiply it by $\frac{\csc x-\cot x}{\csc x-\cot x}$ and substitute $u=\csc x-\cot x.$
• Feb 24th 2008, 06:35 PM
Jhevon
Quote:

Originally Posted by Krizalid
Multiply it by $\frac{\csc x-\cot x}{\csc x-\cot x}$ and substitute $u=\csc x-\cot x.$

wow

i would have told him to look it up in an integration table :D
• Feb 24th 2008, 07:25 PM
mr fantastic
Quote:

Originally Posted by Krizalid
Multiply it by $\frac{\csc x-\cot x}{\csc x-\cot x}$ and substitute $u=\csc x-\cot x.$

Or use the usual Weiestrass substitution.
• Feb 24th 2008, 08:40 PM
Peritus
or:

$\begin{gathered}
\int {\frac{1}
{{\sin x}}dx} = \int {\frac{{\sin x}}
{{\sin ^2 x}}dx} \hfill \\
\hfill \\
t = \cos x \hfill \\
dt = - \sin xdx \hfill \\
\end{gathered}$

$
- \int {\frac{1}
{{1 - t^2 }}dt = } - \int {\frac{1}
{{1 - t^2 }}dt = } - \frac{1}
{2}\int {\frac{1}
{{1 - t}} + \frac{1}
{{1 + t}}} dt = \frac{1}
{2}\left[ {\ln \left( {1 - t} \right) - \ln (1 + t)} \right] =
$

$
= \frac{1}
{2}\ln \left( {\frac{{1 - t}}
{{1 + t}}} \right) = \ln \sqrt {\frac{{1 - \cos x}}
{{1 + \cos x}}}

$
• Mar 4th 2008, 09:30 AM
eng.sherif
hay i have the solution after simplfying u will reach to(let S is integral) S(cosx+cscx)dx=
S(cosx)dx+S(cscx)dx=
so the proplem to integrate csx look multiply by csx up and down u will reach to csx^2x/cscx remove the down one and write root of(1+cotan^2x) and the add negative to the uppers cotan^2x and a negatibe out of integral so u will have the result sinh-1(cotanx)
so the final answe is sinx+sinh-1(cotanx)+c and if u need help i have 4 ways for integrating cscx my mail is sherif.el-sayed@student.guc.edu.eg