How to evaluate the definite integral $int_0^a/2 x^2(a^2-x^2)^-3/2dx $

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$$int_0^a/2x^2(a^2-x^2)^-3/2dx $$




I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.




integration trigonometry substitution




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    $$int_0^a/2x^2(a^2-x^2)^-3/2dx $$




    I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
    When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
    I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.




    integration trigonometry substitution




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      $$int_0^a/2x^2(a^2-x^2)^-3/2dx $$




      I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
      When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
      I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.




      integration trigonometry substitution




      share|cite|improve this question














      $$int_0^a/2x^2(a^2-x^2)^-3/2dx $$




      I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
      When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
      I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.





      integration trigonometry substitution





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Aug 3 at 10:34









      Dylan

      11.4k31026




      11.4k31026









      asked Aug 2 at 3:49









      Maiz Halyym

      61




      61




















          2 Answers
          2






          active

          oldest

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          up vote
          2
          down vote













          Hint: Using substitution $x=asin t$ then
          $$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$






          share|cite|improve this answer





















          • That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
            – Maiz Halyym
            Aug 2 at 4:09











          • I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
            – user 108128
            Aug 2 at 4:13











          • I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
            – Maiz Halyym
            Aug 2 at 4:20










          • What's te final answer ?
            – user 108128
            Aug 2 at 4:21










          • What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
            – Maiz Halyym
            Aug 2 at 4:31

















          up vote
          0
          down vote













          Hint:



          How about integration by parts



          $$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$



          Now choose $sqrta^2-x^2=y$






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote













            Hint: Using substitution $x=asin t$ then
            $$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$






            share|cite|improve this answer





















            • That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
              – Maiz Halyym
              Aug 2 at 4:09











            • I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
              – user 108128
              Aug 2 at 4:13











            • I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
              – Maiz Halyym
              Aug 2 at 4:20










            • What's te final answer ?
              – user 108128
              Aug 2 at 4:21










            • What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
              – Maiz Halyym
              Aug 2 at 4:31














            up vote
            2
            down vote













            Hint: Using substitution $x=asin t$ then
            $$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$






            share|cite|improve this answer





















            • That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
              – Maiz Halyym
              Aug 2 at 4:09











            • I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
              – user 108128
              Aug 2 at 4:13











            • I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
              – Maiz Halyym
              Aug 2 at 4:20










            • What's te final answer ?
              – user 108128
              Aug 2 at 4:21










            • What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
              – Maiz Halyym
              Aug 2 at 4:31












            up vote
            2
            down vote










            up vote
            2
            down vote









            Hint: Using substitution $x=asin t$ then
            $$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$






            share|cite|improve this answer













            Hint: Using substitution $x=asin t$ then
            $$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Aug 2 at 4:05









            user 108128

            18.8k41544




            18.8k41544











            • That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
              – Maiz Halyym
              Aug 2 at 4:09











            • I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
              – user 108128
              Aug 2 at 4:13











            • I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
              – Maiz Halyym
              Aug 2 at 4:20










            • What's te final answer ?
              – user 108128
              Aug 2 at 4:21










            • What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
              – Maiz Halyym
              Aug 2 at 4:31
















            • That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
              – Maiz Halyym
              Aug 2 at 4:09











            • I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
              – user 108128
              Aug 2 at 4:13











            • I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
              – Maiz Halyym
              Aug 2 at 4:20










            • What's te final answer ?
              – user 108128
              Aug 2 at 4:21










            • What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
              – Maiz Halyym
              Aug 2 at 4:31















            That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
            – Maiz Halyym
            Aug 2 at 4:09





            That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
            – Maiz Halyym
            Aug 2 at 4:09













            I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
            – user 108128
            Aug 2 at 4:13





            I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
            – user 108128
            Aug 2 at 4:13













            I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
            – Maiz Halyym
            Aug 2 at 4:20




            I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
            – Maiz Halyym
            Aug 2 at 4:20












            What's te final answer ?
            – user 108128
            Aug 2 at 4:21




            What's te final answer ?
            – user 108128
            Aug 2 at 4:21












            What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
            – Maiz Halyym
            Aug 2 at 4:31




            What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
            – Maiz Halyym
            Aug 2 at 4:31










            up vote
            0
            down vote













            Hint:



            How about integration by parts



            $$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$



            Now choose $sqrta^2-x^2=y$






            share|cite|improve this answer

























              up vote
              0
              down vote













              Hint:



              How about integration by parts



              $$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$



              Now choose $sqrta^2-x^2=y$






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                Hint:



                How about integration by parts



                $$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$



                Now choose $sqrta^2-x^2=y$






                share|cite|improve this answer













                Hint:



                How about integration by parts



                $$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$



                Now choose $sqrta^2-x^2=y$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Aug 2 at 9:40









                lab bhattacharjee

                214k14152263




                214k14152263






















                     

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