This discussion is now closed.

Check out other Related discussions

- AQA A-level Further Maths Required Trig Functions?
- STEP II/III 2014 solutions
- STEP Maths I, II, III 1993 Solutions
- Ball Dropped Onto Spring
- STEP Maths I, II, III 1989 solutions
- differentiation help
- STEP Maths I, II, III 2001 Solutions
- STEP Maths I,II,III 1987 Solutions
- Complex Numbers Question
- Calculator not giving exact answers
- OCR A Level Mathematics B (MEI) Paper 1 (H640/01) - 4th June 2024 [Exam Chat]
- AQA A-level Further Mathematics Paper 1 (7367/1) - 22nd May 2024 [Exam Chat]
- MEI A Level Further Maths Core Pure 22nd May 2024
- maths help!
- Dịch vụ môi trường uy tín là đơn vị hoạt động chuyên nghiệp trong lĩnh vực môi trường
- STEP 2006 Solutions Thread
- Implicit Differentiation 2
- Trigonometric identities
- mirror equation
- reflection about y=x+2

I can think of a way, but it's very dirty.

Consider

I = $\int$ (2 e^x)/(1 + e^(2x)) dx

This integrates to 2arctan(e^x) + C_1.

On the other hand, we can re-write I as:

$\int$ 2/(e^(-x) + e^x) dx

= $\int$ 1/cosh(x) dx

= $\int$ cosh(x)/cosh^2(x) dx

= $\int$ cosh(x)/(1 + sinh^2(x)) dx

= $\int$ 1/(1 + u^2) du ---- (let u = sinh(x))

= arctan(u) + C_2

= arctan(sinh(x)) + C_2

Thus:

arctan(sinh(x)) = 2arctan(e^x) + C --- (where C = C_1 - C_2 is constant)

Now let x=0, and we find that:

0 = 2 * pi/4 + C

=> C = -pi/2

That is,

arctan(sinh(x)) = 2arctan(e^x) - pi/2

I think I'm gonna go take a shower now.

Consider

I = $\int$ (2 e^x)/(1 + e^(2x)) dx

This integrates to 2arctan(e^x) + C_1.

On the other hand, we can re-write I as:

$\int$ 2/(e^(-x) + e^x) dx

= $\int$ 1/cosh(x) dx

= $\int$ cosh(x)/cosh^2(x) dx

= $\int$ cosh(x)/(1 + sinh^2(x)) dx

= $\int$ 1/(1 + u^2) du ---- (let u = sinh(x))

= arctan(u) + C_2

= arctan(sinh(x)) + C_2

Thus:

arctan(sinh(x)) = 2arctan(e^x) + C --- (where C = C_1 - C_2 is constant)

Now let x=0, and we find that:

0 = 2 * pi/4 + C

=> C = -pi/2

That is,

arctan(sinh(x)) = 2arctan(e^x) - pi/2

I think I'm gonna go take a shower now.

Maybe I'm missing something, but I think this is not so bad.

Put tan A = e^x, tan B = e^-x.

Then $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} = \frac{e^x-e^{-x}}{2} = \sinh x$

So $A - B = \tan^{-1}(\sinh x)$.

But $A = \tan^{-1}(e^x), B = \pi/2 - \tan^{-1}(e^x)$.

So $2 \tan^{-1}(e^x) - \pi/2 = \tan^{-1}(\sinh x)$.

Put tan A = e^x, tan B = e^-x.

Then $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} = \frac{e^x-e^{-x}}{2} = \sinh x$

So $A - B = \tan^{-1}(\sinh x)$.

But $A = \tan^{-1}(e^x), B = \pi/2 - \tan^{-1}(e^x)$.

So $2 \tan^{-1}(e^x) - \pi/2 = \tan^{-1}(\sinh x)$.

DFranklin

Maybe I'm missing something, but I think this is not so bad.

Put tan A = e^x, tan B = e^-x.

Then $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} = \frac{e^x-e^{-x}}{2} = \sinh x$

So $A - B = \tan^{-1}(\sinh x)$.

But $A = \tan^{-1}(e^x), B = \pi/2 - \tan^{-1}(e^x)$.

So $2 \tan^{-1}(e^x) - \pi/2 = \tan^{-1}(\sinh x)$.

Put tan A = e^x, tan B = e^-x.

Then $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} = \frac{e^x-e^{-x}}{2} = \sinh x$

So $A - B = \tan^{-1}(\sinh x)$.

But $A = \tan^{-1}(e^x), B = \pi/2 - \tan^{-1}(e^x)$.

So $2 \tan^{-1}(e^x) - \pi/2 = \tan^{-1}(\sinh x)$.

Thanks a bunch... you're right, it's not so bad, in fact it's a bit nice. I was thinking along this line, but God knows why I only thought of tan(A+B) and not tan(A-B)

- AQA A-level Further Maths Required Trig Functions?
- STEP II/III 2014 solutions
- STEP Maths I, II, III 1993 Solutions
- Ball Dropped Onto Spring
- STEP Maths I, II, III 1989 solutions
- differentiation help
- STEP Maths I, II, III 2001 Solutions
- STEP Maths I,II,III 1987 Solutions
- Complex Numbers Question
- Calculator not giving exact answers
- OCR A Level Mathematics B (MEI) Paper 1 (H640/01) - 4th June 2024 [Exam Chat]
- AQA A-level Further Mathematics Paper 1 (7367/1) - 22nd May 2024 [Exam Chat]
- MEI A Level Further Maths Core Pure 22nd May 2024
- maths help!
- Dịch vụ môi trường uy tín là đơn vị hoạt động chuyên nghiệp trong lĩnh vực môi trường
- STEP 2006 Solutions Thread
- Implicit Differentiation 2
- Trigonometric identities
- mirror equation
- reflection about y=x+2

Latest

Trending

Last reply 1 month ago

How do l find the min & max radius of a circle on an argand diagramMaths

2

4