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Finding $int_-pi /2^pi /2 fraclog (1 + b sin x)sin x,mathrm dx$ given $|b|

Proving elementary, $int_0^2pilog frac(1+sin x)^1+cos x1+cos x mathrmdx=0$Integral $I:=int_0^1 fraclog^2 xx^2-x+1mathrm dx=frac10pi^381 sqrt 3$Finding $displaystyle int_1^infty fracsin^4(log x)x^2 log x mathrmdx$Evaluate $int_0^inftyfraclog^2 xe^x^2mathrmdx$.How do you solve $int_0^1int_e^x^efracdydxlogy$?Evaluate $int_-1^1fracsin(x)arcsin(x)}dx$Show that $intlimits_0^{fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$How to evaluate $int_-infty^inftyfracxarctanfrac1x log(1+x^2)1+x^2dx$Evaluate $int_-pi/4^pi/4fracxsin xmathrmdx$Calculate the integral: $int_0^1 |sin log x| ,mathrmdx$

3

1

$begingroup$

Find $$int_-pi/2^pi/2fraclog(1+bsin x)sin x,mathrm dx$$given that $|b|<1$.

I split the integral into$$I=int_0^pi/2f(x),mathrm dx+int_-pi/2^0f(x),mathrm dx$$
For the second term made the substitution $x =-t$ and further solved $I$ to get$$I=int_0 ^pi / 2 fraclog frac1 + b sin x1- b sin xsin x,mathrm dx$$

I do not know how to proceed further. The answer is $pi arcsin b$.

definite-integrals

share|cite|improve this question

edited Mar 15 at 3:35

Saad

20.1k92352

asked Mar 15 at 3:18

CaptainQuestionCaptainQuestion

1769

$endgroup$

add a comment | 

3

1

$begingroup$

Find $$int_-pi/2^pi/2fraclog(1+bsin x)sin x,mathrm dx$$given that $|b|<1$.

I split the integral into$$I=int_0^pi/2f(x),mathrm dx+int_-pi/2^0f(x),mathrm dx$$
For the second term made the substitution $x =-t$ and further solved $I$ to get$$I=int_0 ^pi / 2 fraclog frac1 + b sin x1- b sin xsin x,mathrm dx$$

I do not know how to proceed further. The answer is $pi arcsin b$.

definite-integrals

share|cite|improve this question

edited Mar 15 at 3:35

Saad

20.1k92352

asked Mar 15 at 3:18

CaptainQuestionCaptainQuestion

1769

$endgroup$

add a comment | 

$begingroup$

Find $$int_-pi/2^pi/2fraclog(1+bsin x)sin x,mathrm dx$$given that $|b|<1$.

I split the integral into$$I=int_0^pi/2f(x),mathrm dx+int_-pi/2^0f(x),mathrm dx$$
For the second term made the substitution $x =-t$ and further solved $I$ to get$$I=int_0 ^pi / 2 fraclog frac1 + b sin x1- b sin xsin x,mathrm dx$$

I do not know how to proceed further. The answer is $pi arcsin b$.

definite-integrals

share|cite|improve this question

edited Mar 15 at 3:35

Saad

20.1k92352

asked Mar 15 at 3:18

CaptainQuestionCaptainQuestion

1769

$endgroup$

share|cite|improve this question

edited Mar 15 at 3:35

Saad

20.1k92352

asked Mar 15 at 3:18

CaptainQuestionCaptainQuestion

1769

share|cite|improve this question

edited Mar 15 at 3:35

Saad

20.1k92352

asked Mar 15 at 3:18

CaptainQuestionCaptainQuestion

1769

edited Mar 15 at 3:35

Saad

20.1k92352

edited Mar 15 at 3:35

Saad

20.1k92352

asked Mar 15 at 3:18

CaptainQuestionCaptainQuestion

1769

asked Mar 15 at 3:18

CaptainQuestionCaptainQuestion

1769

1 Answer
1

active

oldest

votes

7

$begingroup$

Let $I(b)=int_-pi/2^pi/2fraclog(1+bsin(x))sin(x),dx$. Differentiating reveals

$$beginalign
I'(b)&=int_-pi/2^pi/2 frac11+bsin(x),dx\\
&=2left(fracarctanleft(sqrtfrac1+b1-bright)+arctanleft(sqrtfrac1-b1+bright)sqrt1-b^2right)\\
&=fracpisqrt1-b^2tag1
endalign$$

Using $I(0)=0$ and integrating $(1)$ yields

$$I(b)=piarcsin(b)$$

share|cite|improve this answer

edited Mar 15 at 3:40

answered Mar 15 at 3:35

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are you allowed to differentiate inside the integral?? how did you get the first step
  $endgroup$
  – terrace
  Mar 15 at 3:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @terrace en.wikipedia.org/wiki/Leibniz_integral_rule
  $endgroup$
  – CaptainQuestion
  Mar 15 at 3:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CaptainQuestion thank you for an interesting link, +1
  $endgroup$
  – Zubin Mukerjee
  Mar 15 at 3:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @terrace Yes, differentiating under the integral is legitimate here.
  $endgroup$
  – Mark Viola
  Mar 15 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Once again, the beautiful trick !
  $endgroup$
  – Claude Leibovici
  Mar 15 at 4:44
  

  
  
  
  

 | 
show 1 more comment

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7

$begingroup$

Let $I(b)=int_-pi/2^pi/2fraclog(1+bsin(x))sin(x),dx$. Differentiating reveals

$$beginalign
I'(b)&=int_-pi/2^pi/2 frac11+bsin(x),dx\\
&=2left(fracarctanleft(sqrtfrac1+b1-bright)+arctanleft(sqrtfrac1-b1+bright)sqrt1-b^2right)\\
&=fracpisqrt1-b^2tag1
endalign$$

Using $I(0)=0$ and integrating $(1)$ yields

$$I(b)=piarcsin(b)$$

share|cite|improve this answer

edited Mar 15 at 3:40

answered Mar 15 at 3:35

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are you allowed to differentiate inside the integral?? how did you get the first step
  $endgroup$
  – terrace
  Mar 15 at 3:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @terrace en.wikipedia.org/wiki/Leibniz_integral_rule
  $endgroup$
  – CaptainQuestion
  Mar 15 at 3:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CaptainQuestion thank you for an interesting link, +1
  $endgroup$
  – Zubin Mukerjee
  Mar 15 at 3:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @terrace Yes, differentiating under the integral is legitimate here.
  $endgroup$
  – Mark Viola
  Mar 15 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Once again, the beautiful trick !
  $endgroup$
  – Claude Leibovici
  Mar 15 at 4:44
  

  
  
  
  

 | 
show 1 more comment

7

$begingroup$

Let $I(b)=int_-pi/2^pi/2fraclog(1+bsin(x))sin(x),dx$. Differentiating reveals

$$beginalign
I'(b)&=int_-pi/2^pi/2 frac11+bsin(x),dx\\
&=2left(fracarctanleft(sqrtfrac1+b1-bright)+arctanleft(sqrtfrac1-b1+bright)sqrt1-b^2right)\\
&=fracpisqrt1-b^2tag1
endalign$$

Using $I(0)=0$ and integrating $(1)$ yields

$$I(b)=piarcsin(b)$$

share|cite|improve this answer

edited Mar 15 at 3:40

answered Mar 15 at 3:35

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are you allowed to differentiate inside the integral?? how did you get the first step
  $endgroup$
  – terrace
  Mar 15 at 3:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @terrace en.wikipedia.org/wiki/Leibniz_integral_rule
  $endgroup$
  – CaptainQuestion
  Mar 15 at 3:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CaptainQuestion thank you for an interesting link, +1
  $endgroup$
  – Zubin Mukerjee
  Mar 15 at 3:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @terrace Yes, differentiating under the integral is legitimate here.
  $endgroup$
  – Mark Viola
  Mar 15 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Once again, the beautiful trick !
  $endgroup$
  – Claude Leibovici
  Mar 15 at 4:44
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Let $I(b)=int_-pi/2^pi/2fraclog(1+bsin(x))sin(x),dx$. Differentiating reveals

$$beginalign
I'(b)&=int_-pi/2^pi/2 frac11+bsin(x),dx\\
&=2left(fracarctanleft(sqrtfrac1+b1-bright)+arctanleft(sqrtfrac1-b1+bright)sqrt1-b^2right)\\
&=fracpisqrt1-b^2tag1
endalign$$

Using $I(0)=0$ and integrating $(1)$ yields

$$I(b)=piarcsin(b)$$

share|cite|improve this answer

edited Mar 15 at 3:40

answered Mar 15 at 3:35

Mark ViolaMark Viola

134k1278176

$endgroup$

share|cite|improve this answer

edited Mar 15 at 3:40

answered Mar 15 at 3:35

Mark ViolaMark Viola

134k1278176

answered Mar 15 at 3:35

Mark ViolaMark Viola

134k1278176

answered Mar 15 at 3:35

Mark ViolaMark Viola

134k1278176