Compute $int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$ [duplicate]












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  • Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $

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Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$



So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
Now what?






integration definite-integrals







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marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22


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    0












    $begingroup$



    This question already has an answer here:




    • Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $

      2 answers




    Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$



    So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
    Now what?






    integration definite-integrals







    share|cite|improve this question











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    marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.





















      0












      0








      0





      $begingroup$



      This question already has an answer here:




      • Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $

        2 answers




      Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$



      So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
      Now what?






      integration definite-integrals







      share|cite|improve this question











      $endgroup$





      This question already has an answer here:




      • Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $

        2 answers




      Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$



      So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
      Now what?





      This question already has an answer here:




      • Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $

        2 answers







      integration definite-integrals




      integration definite-integrals






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      edited Jan 24 at 18:52









      Zacky

      7,89011061




      7,89011061










      asked Jan 24 at 18:38









      C. CristiC. Cristi

      1,634218




      1,634218




      marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          1 Answer
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          $begingroup$

          $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
          $$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
          This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
          In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
          $$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
          $$Rightarrow I(y)=piln(1+y) -pi ln 2$$






          share|cite|improve this answer











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          • $begingroup$
            $I = pi ln(1+y)-frac {pi}4 ln 2?$
            $endgroup$
            – C. Cristi
            Jan 24 at 19:21












          • $begingroup$
            Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
            $endgroup$
            – C. Cristi
            Jan 26 at 8:28










          • $begingroup$
            @C.Cristi I have updated the answer.
            $endgroup$
            – Zacky
            Jan 26 at 9:58


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
          $$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
          This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
          In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
          $$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
          $$Rightarrow I(y)=piln(1+y) -pi ln 2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $I = pi ln(1+y)-frac {pi}4 ln 2?$
            $endgroup$
            – C. Cristi
            Jan 24 at 19:21












          • $begingroup$
            Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
            $endgroup$
            – C. Cristi
            Jan 26 at 8:28










          • $begingroup$
            @C.Cristi I have updated the answer.
            $endgroup$
            – Zacky
            Jan 26 at 9:58
















          3












          $begingroup$

          $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
          $$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
          This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
          In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
          $$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
          $$Rightarrow I(y)=piln(1+y) -pi ln 2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $I = pi ln(1+y)-frac {pi}4 ln 2?$
            $endgroup$
            – C. Cristi
            Jan 24 at 19:21












          • $begingroup$
            Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
            $endgroup$
            – C. Cristi
            Jan 26 at 8:28










          • $begingroup$
            @C.Cristi I have updated the answer.
            $endgroup$
            – Zacky
            Jan 26 at 9:58














          3












          3








          3





          $begingroup$

          $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
          $$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
          This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
          In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
          $$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
          $$Rightarrow I(y)=piln(1+y) -pi ln 2$$






          share|cite|improve this answer











          $endgroup$



          $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
          $$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
          $$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
          This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
          In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
          $$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
          $$Rightarrow I(y)=piln(1+y) -pi ln 2$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 26 at 9:57

























          answered Jan 24 at 18:49









          ZackyZacky

          7,89011061




          7,89011061












          • $begingroup$
            $I = pi ln(1+y)-frac {pi}4 ln 2?$
            $endgroup$
            – C. Cristi
            Jan 24 at 19:21












          • $begingroup$
            Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
            $endgroup$
            – C. Cristi
            Jan 26 at 8:28










          • $begingroup$
            @C.Cristi I have updated the answer.
            $endgroup$
            – Zacky
            Jan 26 at 9:58


















          • $begingroup$
            $I = pi ln(1+y)-frac {pi}4 ln 2?$
            $endgroup$
            – C. Cristi
            Jan 24 at 19:21












          • $begingroup$
            Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
            $endgroup$
            – C. Cristi
            Jan 26 at 8:28










          • $begingroup$
            @C.Cristi I have updated the answer.
            $endgroup$
            – Zacky
            Jan 26 at 9:58
















          $begingroup$
          $I = pi ln(1+y)-frac {pi}4 ln 2?$
          $endgroup$
          – C. Cristi
          Jan 24 at 19:21






          $begingroup$
          $I = pi ln(1+y)-frac {pi}4 ln 2?$
          $endgroup$
          – C. Cristi
          Jan 24 at 19:21














          $begingroup$
          Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
          $endgroup$
          – C. Cristi
          Jan 26 at 8:28




          $begingroup$
          Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
          $endgroup$
          – C. Cristi
          Jan 26 at 8:28












          $begingroup$
          @C.Cristi I have updated the answer.
          $endgroup$
          – Zacky
          Jan 26 at 9:58




          $begingroup$
          @C.Cristi I have updated the answer.
          $endgroup$
          – Zacky
          Jan 26 at 9:58



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