Evaluation of Irrational equations involving $2$ square root

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Finding value of $displaystyle int sqrtx+sqrtkx-x^2dx$




My Try: Let $displaystyle I = sqrtx+sqrtkx-x^2$. Substitute $kx-x^2=t^2$



Then $(k-2x)dx = dt$



$Rightarrow displaystyle I = int sqrtx+t; times frac1k-2xdt$



Now $displaystyle x^2-kx+t^2=0Rightarrow x=frackpm sqrtk^2-4t^22$



$displaystyle Rightarrow I = pm int sqrtfrackpm sqrtk^2-4t^22+ttimes frac1sqrtk^2-4t^2dt$



Could some help me how to solve it



please help me , Thanks










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  • 1




    Use $$x-dfrack2=dfrack2sintheta$$
    – Nosrati
    Sep 10 at 8:46










  • It seems that it would end to an Elliptic integral!
    – mrs
    Sep 10 at 8:50






  • 1




    A few steps: With Nosrati's hint $x=dfrack2+dfrack2sintheta$: $$ I = int sqrtx+sqrtkx-x^2dx = \ left[dfrack2right]^3/2 int sqrt1+sintheta +costheta ; costheta ; dtheta \ = left[dfrack2right]^3/2 int sqrtsqrt2 sin(theta + pi/4) + 1 ; costheta ; dtheta \ $$ .....
    – Andreas
    Sep 10 at 12:21






  • 1




    ... Since $$costheta = cos(theta + pi/4 - pi/4) = cos(theta + pi/4) cos(pi/4) -sin(theta + pi/4) sin(pi/4) \= frac1sqrt 2cos(theta + pi/4) - frac1sqrt 2sin(theta + pi/4) $$ set $alpha = theta + pi/4$ and obtain $$ I = left[dfrack2right]^3/2 int sqrtsqrt2 sin(alpha) + 1 ; (frac1sqrt 2cos(alpha) - frac1sqrt 2sin(alpha) ) ; dalpha\ $$ The first one is easy, the second one becomes elliptic
    – Andreas
    Sep 10 at 12:21















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0
down vote

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Finding value of $displaystyle int sqrtx+sqrtkx-x^2dx$




My Try: Let $displaystyle I = sqrtx+sqrtkx-x^2$. Substitute $kx-x^2=t^2$



Then $(k-2x)dx = dt$



$Rightarrow displaystyle I = int sqrtx+t; times frac1k-2xdt$



Now $displaystyle x^2-kx+t^2=0Rightarrow x=frackpm sqrtk^2-4t^22$



$displaystyle Rightarrow I = pm int sqrtfrackpm sqrtk^2-4t^22+ttimes frac1sqrtk^2-4t^2dt$



Could some help me how to solve it



please help me , Thanks










share|cite|improve this question

















  • 1




    Use $$x-dfrack2=dfrack2sintheta$$
    – Nosrati
    Sep 10 at 8:46










  • It seems that it would end to an Elliptic integral!
    – mrs
    Sep 10 at 8:50






  • 1




    A few steps: With Nosrati's hint $x=dfrack2+dfrack2sintheta$: $$ I = int sqrtx+sqrtkx-x^2dx = \ left[dfrack2right]^3/2 int sqrt1+sintheta +costheta ; costheta ; dtheta \ = left[dfrack2right]^3/2 int sqrtsqrt2 sin(theta + pi/4) + 1 ; costheta ; dtheta \ $$ .....
    – Andreas
    Sep 10 at 12:21






  • 1




    ... Since $$costheta = cos(theta + pi/4 - pi/4) = cos(theta + pi/4) cos(pi/4) -sin(theta + pi/4) sin(pi/4) \= frac1sqrt 2cos(theta + pi/4) - frac1sqrt 2sin(theta + pi/4) $$ set $alpha = theta + pi/4$ and obtain $$ I = left[dfrack2right]^3/2 int sqrtsqrt2 sin(alpha) + 1 ; (frac1sqrt 2cos(alpha) - frac1sqrt 2sin(alpha) ) ; dalpha\ $$ The first one is easy, the second one becomes elliptic
    – Andreas
    Sep 10 at 12:21













up vote
0
down vote

favorite









up vote
0
down vote

favorite












Finding value of $displaystyle int sqrtx+sqrtkx-x^2dx$




My Try: Let $displaystyle I = sqrtx+sqrtkx-x^2$. Substitute $kx-x^2=t^2$



Then $(k-2x)dx = dt$



$Rightarrow displaystyle I = int sqrtx+t; times frac1k-2xdt$



Now $displaystyle x^2-kx+t^2=0Rightarrow x=frackpm sqrtk^2-4t^22$



$displaystyle Rightarrow I = pm int sqrtfrackpm sqrtk^2-4t^22+ttimes frac1sqrtk^2-4t^2dt$



Could some help me how to solve it



please help me , Thanks










share|cite|improve this question














Finding value of $displaystyle int sqrtx+sqrtkx-x^2dx$




My Try: Let $displaystyle I = sqrtx+sqrtkx-x^2$. Substitute $kx-x^2=t^2$



Then $(k-2x)dx = dt$



$Rightarrow displaystyle I = int sqrtx+t; times frac1k-2xdt$



Now $displaystyle x^2-kx+t^2=0Rightarrow x=frackpm sqrtk^2-4t^22$



$displaystyle Rightarrow I = pm int sqrtfrackpm sqrtk^2-4t^22+ttimes frac1sqrtk^2-4t^2dt$



Could some help me how to solve it



please help me , Thanks







indefinite-integrals






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asked Sep 10 at 8:34









Durgesh Tiwari

4,8152526




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  • 1




    Use $$x-dfrack2=dfrack2sintheta$$
    – Nosrati
    Sep 10 at 8:46










  • It seems that it would end to an Elliptic integral!
    – mrs
    Sep 10 at 8:50






  • 1




    A few steps: With Nosrati's hint $x=dfrack2+dfrack2sintheta$: $$ I = int sqrtx+sqrtkx-x^2dx = \ left[dfrack2right]^3/2 int sqrt1+sintheta +costheta ; costheta ; dtheta \ = left[dfrack2right]^3/2 int sqrtsqrt2 sin(theta + pi/4) + 1 ; costheta ; dtheta \ $$ .....
    – Andreas
    Sep 10 at 12:21






  • 1




    ... Since $$costheta = cos(theta + pi/4 - pi/4) = cos(theta + pi/4) cos(pi/4) -sin(theta + pi/4) sin(pi/4) \= frac1sqrt 2cos(theta + pi/4) - frac1sqrt 2sin(theta + pi/4) $$ set $alpha = theta + pi/4$ and obtain $$ I = left[dfrack2right]^3/2 int sqrtsqrt2 sin(alpha) + 1 ; (frac1sqrt 2cos(alpha) - frac1sqrt 2sin(alpha) ) ; dalpha\ $$ The first one is easy, the second one becomes elliptic
    – Andreas
    Sep 10 at 12:21













  • 1




    Use $$x-dfrack2=dfrack2sintheta$$
    – Nosrati
    Sep 10 at 8:46










  • It seems that it would end to an Elliptic integral!
    – mrs
    Sep 10 at 8:50






  • 1




    A few steps: With Nosrati's hint $x=dfrack2+dfrack2sintheta$: $$ I = int sqrtx+sqrtkx-x^2dx = \ left[dfrack2right]^3/2 int sqrt1+sintheta +costheta ; costheta ; dtheta \ = left[dfrack2right]^3/2 int sqrtsqrt2 sin(theta + pi/4) + 1 ; costheta ; dtheta \ $$ .....
    – Andreas
    Sep 10 at 12:21






  • 1




    ... Since $$costheta = cos(theta + pi/4 - pi/4) = cos(theta + pi/4) cos(pi/4) -sin(theta + pi/4) sin(pi/4) \= frac1sqrt 2cos(theta + pi/4) - frac1sqrt 2sin(theta + pi/4) $$ set $alpha = theta + pi/4$ and obtain $$ I = left[dfrack2right]^3/2 int sqrtsqrt2 sin(alpha) + 1 ; (frac1sqrt 2cos(alpha) - frac1sqrt 2sin(alpha) ) ; dalpha\ $$ The first one is easy, the second one becomes elliptic
    – Andreas
    Sep 10 at 12:21








1




1




Use $$x-dfrack2=dfrack2sintheta$$
– Nosrati
Sep 10 at 8:46




Use $$x-dfrack2=dfrack2sintheta$$
– Nosrati
Sep 10 at 8:46












It seems that it would end to an Elliptic integral!
– mrs
Sep 10 at 8:50




It seems that it would end to an Elliptic integral!
– mrs
Sep 10 at 8:50




1




1




A few steps: With Nosrati's hint $x=dfrack2+dfrack2sintheta$: $$ I = int sqrtx+sqrtkx-x^2dx = \ left[dfrack2right]^3/2 int sqrt1+sintheta +costheta ; costheta ; dtheta \ = left[dfrack2right]^3/2 int sqrtsqrt2 sin(theta + pi/4) + 1 ; costheta ; dtheta \ $$ .....
– Andreas
Sep 10 at 12:21




A few steps: With Nosrati's hint $x=dfrack2+dfrack2sintheta$: $$ I = int sqrtx+sqrtkx-x^2dx = \ left[dfrack2right]^3/2 int sqrt1+sintheta +costheta ; costheta ; dtheta \ = left[dfrack2right]^3/2 int sqrtsqrt2 sin(theta + pi/4) + 1 ; costheta ; dtheta \ $$ .....
– Andreas
Sep 10 at 12:21




1




1




... Since $$costheta = cos(theta + pi/4 - pi/4) = cos(theta + pi/4) cos(pi/4) -sin(theta + pi/4) sin(pi/4) \= frac1sqrt 2cos(theta + pi/4) - frac1sqrt 2sin(theta + pi/4) $$ set $alpha = theta + pi/4$ and obtain $$ I = left[dfrack2right]^3/2 int sqrtsqrt2 sin(alpha) + 1 ; (frac1sqrt 2cos(alpha) - frac1sqrt 2sin(alpha) ) ; dalpha\ $$ The first one is easy, the second one becomes elliptic
– Andreas
Sep 10 at 12:21





... Since $$costheta = cos(theta + pi/4 - pi/4) = cos(theta + pi/4) cos(pi/4) -sin(theta + pi/4) sin(pi/4) \= frac1sqrt 2cos(theta + pi/4) - frac1sqrt 2sin(theta + pi/4) $$ set $alpha = theta + pi/4$ and obtain $$ I = left[dfrack2right]^3/2 int sqrtsqrt2 sin(alpha) + 1 ; (frac1sqrt 2cos(alpha) - frac1sqrt 2sin(alpha) ) ; dalpha\ $$ The first one is easy, the second one becomes elliptic
– Andreas
Sep 10 at 12:21
















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