Evaluation of Irrational equations involving $2$ square root

Multi tool use
Clash Royale CLAN TAG#URR8PPP
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
indefinite-integrals
add a comment |Â
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
indefinite-integrals
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
add a comment |Â
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
indefinite-integrals
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
indefinite-integrals
asked Sep 10 at 8:34
Durgesh Tiwari
4,8152526
4,8152526
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
add a comment |Â
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
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911677%2fevaluation-of-irrational-equations-involving-2-square-root%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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