Integrating (sinx + secx)/tanx dx

**L_U_K_E** · February 24th 2008, 04:25 PM

∫(sinx + secx) / tanx dx

∫(sinx/tanx)dx + ∫(secx/tanx)dx

∫(cosx)dx + ∫(1/sinx)dx

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

**Soroban** · February 24th 2008, 04:55 PM

Hello, L_U_K_E!

Just one more step . . .

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

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

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

Got it?

**L_U_K_E** · February 24th 2008, 04:59 PM

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?

**Krizalid** · February 24th 2008, 05:03 PM

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

**Jhevon** · February 24th 2008, 05:35 PM

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

**mr fantastic** · February 24th 2008, 06:25 PM

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.

**Peritus** · February 24th 2008, 07:40 PM

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}}} $

**eng.sherif** · March 4th 2008, 08:30 AM

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

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