Evaluate Integral $int_{frac{r}{2}}^{r}(e^{2t-2r}- e^{-2t})^{n-2} dt$ [closed]

-1

I want to Evaluate Integral

$int_{frac{r}{2}}^{r}(e^{2t-2r}- e^{-2t})^{n-2} dt$

the well known way is to expand and then calculate the integral, but I am really interested finding another solutions

Any help is appreciated!

asked Dec 12 '18 at 13:25

user62498

1,948614

closed as off-topic by Nosrati, Saad, Ben, Cesareo, José Carlos Santos Dec 14 '18 at 10:27

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"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, Ben, Cesareo, José Carlos Santos

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add a comment |

-1

I want to Evaluate Integral

$int_{frac{r}{2}}^{r}(e^{2t-2r}- e^{-2t})^{n-2} dt$

the well known way is to expand and then calculate the integral, but I am really interested finding another solutions

Any help is appreciated!

asked Dec 12 '18 at 13:25

user62498

1,948614

closed as off-topic by Nosrati, Saad, Ben, Cesareo, José Carlos Santos Dec 14 '18 at 10:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, Ben, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

-1

I want to Evaluate Integral

$int_{frac{r}{2}}^{r}(e^{2t-2r}- e^{-2t})^{n-2} dt$

the well known way is to expand and then calculate the integral, but I am really interested finding another solutions

Any help is appreciated!

asked Dec 12 '18 at 13:25

user62498

1,948614

I want to Evaluate Integral

$int_{frac{r}{2}}^{r}(e^{2t-2r}- e^{-2t})^{n-2} dt$

the well known way is to expand and then calculate the integral, but I am really interested finding another solutions

Any help is appreciated!

calculus integration

asked Dec 12 '18 at 13:25

user62498

1,948614

asked Dec 12 '18 at 13:25

user62498

1,948614

asked Dec 12 '18 at 13:25

user62498

1,948614

asked Dec 12 '18 at 13:25

user62498

1,948614

asked Dec 12 '18 at 13:25

user62498

1,948614

closed as off-topic by Nosrati, Saad, Ben, Cesareo, José Carlos Santos Dec 14 '18 at 10:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, Ben, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Nosrati, Saad, Ben, Cesareo, José Carlos Santos Dec 14 '18 at 10:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, Ben, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

1 Answer
1

active

oldest

votes

Hint.

Changing variables $y = t-frac r2$ we have $dt = dy$ and

$$
e^{2t-2r} - e^{-2t} = 4e^{-r}cosh ysinh y = 2sinh(2y)e^{-r}
$$

then after that making $z = 2y$

$$
frac{d}{dz}left(sinh^n zcosh zright) = (n+1)sinh^{n+1}z + n sinh ^{n-1}z
$$

now we can construct the recurrence formula

$$
(n+1)I_{n+1}+n I_{n-1} = sinh^n zcosh z|_a^b
$$

with

$$
I_n = int_a^b sinh^n z dz
$$

edited Dec 12 '18 at 15:52

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

$begingroup$
how can this be helpful
$endgroup$
– Nosrati
Dec 12 '18 at 13:44

$begingroup$
@Cesareo, Thanks, I guess we have problem to find integral $sinh^{n-2}(2y)$
$endgroup$
– user62498
Dec 12 '18 at 13:53

$begingroup$
@user62498 See the attached recurrence formula.
$endgroup$
– Cesareo
Dec 12 '18 at 14:02

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Hint.

Changing variables $y = t-frac r2$ we have $dt = dy$ and

$$
e^{2t-2r} - e^{-2t} = 4e^{-r}cosh ysinh y = 2sinh(2y)e^{-r}
$$

then after that making $z = 2y$

$$
frac{d}{dz}left(sinh^n zcosh zright) = (n+1)sinh^{n+1}z + n sinh ^{n-1}z
$$

now we can construct the recurrence formula

$$
(n+1)I_{n+1}+n I_{n-1} = sinh^n zcosh z|_a^b
$$

with

$$
I_n = int_a^b sinh^n z dz
$$

edited Dec 12 '18 at 15:52

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

$begingroup$
how can this be helpful
$endgroup$
– Nosrati
Dec 12 '18 at 13:44

$begingroup$
@Cesareo, Thanks, I guess we have problem to find integral $sinh^{n-2}(2y)$
$endgroup$
– user62498
Dec 12 '18 at 13:53

$begingroup$
@user62498 See the attached recurrence formula.
$endgroup$
– Cesareo
Dec 12 '18 at 14:02

add a comment |

Hint.

Changing variables $y = t-frac r2$ we have $dt = dy$ and

$$
e^{2t-2r} - e^{-2t} = 4e^{-r}cosh ysinh y = 2sinh(2y)e^{-r}
$$

then after that making $z = 2y$

$$
frac{d}{dz}left(sinh^n zcosh zright) = (n+1)sinh^{n+1}z + n sinh ^{n-1}z
$$

now we can construct the recurrence formula

$$
(n+1)I_{n+1}+n I_{n-1} = sinh^n zcosh z|_a^b
$$

with

$$
I_n = int_a^b sinh^n z dz
$$

edited Dec 12 '18 at 15:52

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

$begingroup$
how can this be helpful
$endgroup$
– Nosrati
Dec 12 '18 at 13:44

$begingroup$
@Cesareo, Thanks, I guess we have problem to find integral $sinh^{n-2}(2y)$
$endgroup$
– user62498
Dec 12 '18 at 13:53

$begingroup$
@user62498 See the attached recurrence formula.
$endgroup$
– Cesareo
Dec 12 '18 at 14:02

add a comment |

Hint.

Changing variables $y = t-frac r2$ we have $dt = dy$ and

$$
e^{2t-2r} - e^{-2t} = 4e^{-r}cosh ysinh y = 2sinh(2y)e^{-r}
$$

then after that making $z = 2y$

$$
frac{d}{dz}left(sinh^n zcosh zright) = (n+1)sinh^{n+1}z + n sinh ^{n-1}z
$$

now we can construct the recurrence formula

$$
(n+1)I_{n+1}+n I_{n-1} = sinh^n zcosh z|_a^b
$$

with

$$
I_n = int_a^b sinh^n z dz
$$

edited Dec 12 '18 at 15:52

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

Hint.

Changing variables $y = t-frac r2$ we have $dt = dy$ and

$$
e^{2t-2r} - e^{-2t} = 4e^{-r}cosh ysinh y = 2sinh(2y)e^{-r}
$$

then after that making $z = 2y$

$$
frac{d}{dz}left(sinh^n zcosh zright) = (n+1)sinh^{n+1}z + n sinh ^{n-1}z
$$

now we can construct the recurrence formula

$$
(n+1)I_{n+1}+n I_{n-1} = sinh^n zcosh z|_a^b
$$

with

$$
I_n = int_a^b sinh^n z dz
$$

edited Dec 12 '18 at 15:52

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

edited Dec 12 '18 at 15:52

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

answered Dec 12 '18 at 13:39

Cesareo

9,0413516

$begingroup$
how can this be helpful
$endgroup$
– Nosrati
Dec 12 '18 at 13:44

$begingroup$
@Cesareo, Thanks, I guess we have problem to find integral $sinh^{n-2}(2y)$
$endgroup$
– user62498
Dec 12 '18 at 13:53

$begingroup$
@user62498 See the attached recurrence formula.
$endgroup$
– Cesareo
Dec 12 '18 at 14:02

add a comment |

$begingroup$
how can this be helpful
$endgroup$
– Nosrati
Dec 12 '18 at 13:44

$begingroup$
@Cesareo, Thanks, I guess we have problem to find integral $sinh^{n-2}(2y)$
$endgroup$
– user62498
Dec 12 '18 at 13:53

$begingroup$
@user62498 See the attached recurrence formula.
$endgroup$
– Cesareo
Dec 12 '18 at 14:02

how can this be helpful

– Nosrati
Dec 12 '18 at 13:44

@Cesareo, Thanks, I guess we have problem to find integral $sinh^{n-2}(2y)$

– user62498
Dec 12 '18 at 13:53

@user62498 See the attached recurrence formula.

– Cesareo
Dec 12 '18 at 14:02

add a comment |

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