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Evaluating $int sqrtfrac3-x3+x sin^-1(sqrtfrac3-x6)dx$
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2
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Evaluate $$int sqrtfrac3-x3+x sin^-1left(sqrtfrac3-x6right)dx.$$
My attempt: I have used the substitution $x=3cos 2theta$ as it seemed appropriate on seeing the integrand. After simplification i got -$12int theta( sin^2 theta) dtheta $ which i was able to evaluate using integration by parts but i got a bit complicated answer. Was my approach correct? What other more efficient methods are there to evaluate this integral?
calculus integration
asked Jul 27 at 21:32
MrAP
1,17821327
 |Â
show 3 more comments
up vote
2
down vote
favorite
Evaluate $$int sqrtfrac3-x3+x sin^-1left(sqrtfrac3-x6right)dx.$$
My attempt: I have used the substitution $x=3cos 2theta$ as it seemed appropriate on seeing the integrand. After simplification i got -$12int theta( sin^2 theta) dtheta $ which i was able to evaluate using integration by parts but i got a bit complicated answer. Was my approach correct? What other more efficient methods are there to evaluate this integral?
calculus integration
asked Jul 27 at 21:32
MrAP
1,17821327
My mistake, thanks
– Doug M
Jul 27 at 21:46
Maybe helpful $$ frac 3-x3+x=frac3-x6+(frac3-x6)(frac 3-x3+x) $$
– user0410
Jul 27 at 21:59
Why is $x=3cos 2 theta$ a reasonable substitution?
– JavaMan
Jul 28 at 1:44
@JavaMan, If you put $x=3cos 2theta$, then, $3-x=3-3cos 2theta=3(1-cos 2theta)=6sin^2 theta$ and $3+x=3+3cos 2theta=3(1+cos 2theta)=6cos^2 theta$. Thus the expression inside the first square root becomes $frac6sin^2theta6cos^2theta$ which results in $tan^2theta$ and due to the square root, the result is $tantheta$.
– MrAP
Jul 29 at 11:36
The second square root simplifies to $sintheta$ and $dx=-6sin 2theta$. Thus, after complete substitution the integral becomes $int (tantheta) (sin^-1sintheta) (-6sin 2theta)dtheta=-12inttheta (sin^2theta) dtheta$.
– MrAP
Jul 29 at 11:43
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Evaluate $$int sqrtfrac3-x3+x sin^-1left(sqrtfrac3-x6right)dx.$$
My attempt: I have used the substitution $x=3cos 2theta$ as it seemed appropriate on seeing the integrand. After simplification i got -$12int theta( sin^2 theta) dtheta $ which i was able to evaluate using integration by parts but i got a bit complicated answer. Was my approach correct? What other more efficient methods are there to evaluate this integral?
calculus integration
asked Jul 27 at 21:32
MrAP
1,17821327
Evaluate $$int sqrtfrac3-x3+x sin^-1left(sqrtfrac3-x6right)dx.$$
My attempt: I have used the substitution $x=3cos 2theta$ as it seemed appropriate on seeing the integrand. After simplification i got -$12int theta( sin^2 theta) dtheta $ which i was able to evaluate using integration by parts but i got a bit complicated answer. Was my approach correct? What other more efficient methods are there to evaluate this integral?
calculus integration
asked Jul 27 at 21:32
MrAP
1,17821327
edited Jul 29 at 12:55
asked Jul 27 at 21:32
MrAP
1,17821327
asked Jul 27 at 21:32
MrAP
1,17821327
asked Jul 27 at 21:32
MrAP
1,17821327
1,17821327
My mistake, thanks
– Doug M
Jul 27 at 21:46
Maybe helpful $$ frac 3-x3+x=frac3-x6+(frac3-x6)(frac 3-x3+x) $$
– user0410
Jul 27 at 21:59
Why is $x=3cos 2 theta$ a reasonable substitution?
– JavaMan
Jul 28 at 1:44
@JavaMan, If you put $x=3cos 2theta$, then, $3-x=3-3cos 2theta=3(1-cos 2theta)=6sin^2 theta$ and $3+x=3+3cos 2theta=3(1+cos 2theta)=6cos^2 theta$. Thus the expression inside the first square root becomes $frac6sin^2theta6cos^2theta$ which results in $tan^2theta$ and due to the square root, the result is $tantheta$.
– MrAP
Jul 29 at 11:36
The second square root simplifies to $sintheta$ and $dx=-6sin 2theta$. Thus, after complete substitution the integral becomes $int (tantheta) (sin^-1sintheta) (-6sin 2theta)dtheta=-12inttheta (sin^2theta) dtheta$.
– MrAP
Jul 29 at 11:43
 |Â
show 3 more comments
My mistake, thanks
– Doug M
Jul 27 at 21:46
Maybe helpful $$ frac 3-x3+x=frac3-x6+(frac3-x6)(frac 3-x3+x) $$
– user0410
Jul 27 at 21:59
Why is $x=3cos 2 theta$ a reasonable substitution?
– JavaMan
Jul 28 at 1:44
@JavaMan, If you put $x=3cos 2theta$, then, $3-x=3-3cos 2theta=3(1-cos 2theta)=6sin^2 theta$ and $3+x=3+3cos 2theta=3(1+cos 2theta)=6cos^2 theta$. Thus the expression inside the first square root becomes $frac6sin^2theta6cos^2theta$ which results in $tan^2theta$ and due to the square root, the result is $tantheta$.
– MrAP
Jul 29 at 11:36
The second square root simplifies to $sintheta$ and $dx=-6sin 2theta$. Thus, after complete substitution the integral becomes $int (tantheta) (sin^-1sintheta) (-6sin 2theta)dtheta=-12inttheta (sin^2theta) dtheta$.
– MrAP
Jul 29 at 11:43
My mistake, thanks
– Doug M
Jul 27 at 21:46
My mistake, thanks
– Doug M
Jul 27 at 21:46
Maybe helpful $$ frac 3-x3+x=frac3-x6+(frac3-x6)(frac 3-x3+x) $$
– user0410
Jul 27 at 21:59
Maybe helpful $$ frac 3-x3+x=frac3-x6+(frac3-x6)(frac 3-x3+x) $$
– user0410
Jul 27 at 21:59
Why is $x=3cos 2 theta$ a reasonable substitution?
– JavaMan
Jul 28 at 1:44
Why is $x=3cos 2 theta$ a reasonable substitution?
– JavaMan
Jul 28 at 1:44
@JavaMan, If you put $x=3cos 2theta$, then, $3-x=3-3cos 2theta=3(1-cos 2theta)=6sin^2 theta$ and $3+x=3+3cos 2theta=3(1+cos 2theta)=6cos^2 theta$. Thus the expression inside the first square root becomes $frac6sin^2theta6cos^2theta$ which results in $tan^2theta$ and due to the square root, the result is $tantheta$.
– MrAP
Jul 29 at 11:36
@JavaMan, If you put $x=3cos 2theta$, then, $3-x=3-3cos 2theta=3(1-cos 2theta)=6sin^2 theta$ and $3+x=3+3cos 2theta=3(1+cos 2theta)=6cos^2 theta$. Thus the expression inside the first square root becomes $frac6sin^2theta6cos^2theta$ which results in $tan^2theta$ and due to the square root, the result is $tantheta$.
– MrAP
Jul 29 at 11:36
The second square root simplifies to $sintheta$ and $dx=-6sin 2theta$. Thus, after complete substitution the integral becomes $int (tantheta) (sin^-1sintheta) (-6sin 2theta)dtheta=-12inttheta (sin^2theta) dtheta$.
– MrAP
Jul 29 at 11:43
The second square root simplifies to $sintheta$ and $dx=-6sin 2theta$. Thus, after complete substitution the integral becomes $int (tantheta) (sin^-1sintheta) (-6sin 2theta)dtheta=-12inttheta (sin^2theta) dtheta$.
– MrAP
Jul 29 at 11:43
 |Â
show 3 more comments
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My mistake, thanks
– Doug M
Jul 27 at 21:46
Maybe helpful $$ frac 3-x3+x=frac3-x6+(frac3-x6)(frac 3-x3+x) $$
– user0410
Jul 27 at 21:59
Why is $x=3cos 2 theta$ a reasonable substitution?
– JavaMan
Jul 28 at 1:44
@JavaMan, If you put $x=3cos 2theta$, then, $3-x=3-3cos 2theta=3(1-cos 2theta)=6sin^2 theta$ and $3+x=3+3cos 2theta=3(1+cos 2theta)=6cos^2 theta$. Thus the expression inside the first square root becomes $frac6sin^2theta6cos^2theta$ which results in $tan^2theta$ and due to the square root, the result is $tantheta$.
– MrAP
Jul 29 at 11:36
The second square root simplifies to $sintheta$ and $dx=-6sin 2theta$. Thus, after complete substitution the integral becomes $int (tantheta) (sin^-1sintheta) (-6sin 2theta)dtheta=-12inttheta (sin^2theta) dtheta$.
– MrAP
Jul 29 at 11:43