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Trigonometric and Exponential Integration

positive integer $n$ in definite integrationFinding $sum^infty_n=1int^2(n+1)pi_2npifracxsin x+cos xx^2$Trigonometric exponential IntegrationValue of product of cosines definite integrationArrangement of definite integral in increasing orderFinding $ int^1_-1fracx^3sqrt1-x^2lnbigg(frac1+x1-xbigg)dx$Finding $int^pi_0sin(8x+8sin 3x)dx$If $int^infty_0fracln^2(x)(1-x)^2dx+kint^1_0fracln(1-x)xdx=0.$ Find $k$.Evaluation of inverse of $tan$ integrationWhat is $PQ^-1$ if $P=int^pi_0fracsin(994 x)sin xsin(1332x),dx$ and $Q=int^1_0fracx^338(x^1988-1)x^2-1,dx$?

6

1

$begingroup$

If $displaystyle I_n =int^infty_fracpi2e^-xcos^n(x)dx.$ Then $displaystyle fracI_2018I_2016$ is

Try: using by parts

$$I_n=int^infty_fracpi2e^-xcos^n(x)dx$$

$$I_n=-cos^n(x)cdot e^-xbigg|^infty_fracpi2-nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$

$$I_n=nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$

Could some help me to solve it, Thanks

definite-integrals

share|cite|improve this question

asked Apr 24 at 6:51

DXTDXT

5,9792733

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
  $endgroup$
  – lab bhattacharjee
  Apr 24 at 6:59
  

  
  
  
  

add a comment | 

6

1

$begingroup$

If $displaystyle I_n =int^infty_fracpi2e^-xcos^n(x)dx.$ Then $displaystyle fracI_2018I_2016$ is

Try: using by parts

$$I_n=int^infty_fracpi2e^-xcos^n(x)dx$$

$$I_n=-cos^n(x)cdot e^-xbigg|^infty_fracpi2-nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$

$$I_n=nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$

Could some help me to solve it, Thanks

definite-integrals

share|cite|improve this question

asked Apr 24 at 6:51

DXTDXT

5,9792733

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
  $endgroup$
  – lab bhattacharjee
  Apr 24 at 6:59
  

  
  
  
  

add a comment | 

$begingroup$

If $displaystyle I_n =int^infty_fracpi2e^-xcos^n(x)dx.$ Then $displaystyle fracI_2018I_2016$ is

Try: using by parts

$$I_n=int^infty_fracpi2e^-xcos^n(x)dx$$

$$I_n=-cos^n(x)cdot e^-xbigg|^infty_fracpi2-nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$

$$I_n=nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$

Could some help me to solve it, Thanks

definite-integrals

share|cite|improve this question

asked Apr 24 at 6:51

DXTDXT

5,9792733

$endgroup$

share|cite|improve this question

asked Apr 24 at 6:51

DXTDXT

5,9792733

share|cite|improve this question

asked Apr 24 at 6:51

DXTDXT

5,9792733

asked Apr 24 at 6:51

DXTDXT

5,9792733

asked Apr 24 at 6:51

DXTDXT

5,9792733

2 Answers
2

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oldest

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7

$begingroup$

As you say,

$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$

I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$

share|cite|improve this answer

edited Apr 24 at 11:00

answered Apr 24 at 9:06

Bennett GardinerBennett Gardiner

3,16211540

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  No, the result is $dfrac40703064072325$. (Checked with Alpha.)
  $endgroup$
  – Yves Daoust
  Apr 24 at 10:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I took a typo from OP as my starting point (oops).
  $endgroup$
  – Bennett Gardiner
  Apr 24 at 11:00
  

  
  
  
  

add a comment | 

4

$begingroup$

For brevity we set

$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$

Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$

Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,

$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$

share|cite|improve this answer

edited Apr 24 at 12:56

answered Apr 24 at 9:57

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

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7

$begingroup$

As you say,

$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$

I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$

share|cite|improve this answer

edited Apr 24 at 11:00

answered Apr 24 at 9:06

Bennett GardinerBennett Gardiner

3,16211540

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  No, the result is $dfrac40703064072325$. (Checked with Alpha.)
  $endgroup$
  – Yves Daoust
  Apr 24 at 10:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I took a typo from OP as my starting point (oops).
  $endgroup$
  – Bennett Gardiner
  Apr 24 at 11:00
  

  
  
  
  

add a comment | 

7

$begingroup$

As you say,

$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$

I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$

share|cite|improve this answer

edited Apr 24 at 11:00

answered Apr 24 at 9:06

Bennett GardinerBennett Gardiner

3,16211540

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  No, the result is $dfrac40703064072325$. (Checked with Alpha.)
  $endgroup$
  – Yves Daoust
  Apr 24 at 10:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I took a typo from OP as my starting point (oops).
  $endgroup$
  – Bennett Gardiner
  Apr 24 at 11:00
  

  
  
  
  

add a comment | 

$begingroup$

As you say,

$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$

I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$

share|cite|improve this answer

edited Apr 24 at 11:00

answered Apr 24 at 9:06

Bennett GardinerBennett Gardiner

3,16211540

$endgroup$

share|cite|improve this answer

edited Apr 24 at 11:00

answered Apr 24 at 9:06

Bennett GardinerBennett Gardiner

3,16211540

answered Apr 24 at 9:06

Bennett GardinerBennett Gardiner

3,16211540

answered Apr 24 at 9:06

Bennett GardinerBennett Gardiner

3,16211540

4

$begingroup$

For brevity we set

$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$

Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$

Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,

$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$

share|cite|improve this answer

edited Apr 24 at 12:56

answered Apr 24 at 9:57

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

4

$begingroup$

For brevity we set

$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$

Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$

Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,

$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$

share|cite|improve this answer

edited Apr 24 at 12:56

answered Apr 24 at 9:57

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

$begingroup$

For brevity we set

$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$

Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$

Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,

$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$

share|cite|improve this answer

edited Apr 24 at 12:56

answered Apr 24 at 9:57

Yves DaoustYves Daoust

135k676233

$endgroup$

share|cite|improve this answer

edited Apr 24 at 12:56

answered Apr 24 at 9:57

Yves DaoustYves Daoust

135k676233

answered Apr 24 at 9:57

Yves DaoustYves Daoust

135k676233

answered Apr 24 at 9:57

Yves DaoustYves Daoust

135k676233