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what is the result of this question: sqrt{1-t^4}.2tdt

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\[\int\limits \sqrt{1-t^4}.2tdt\]
\[u=t^2~~~and ~~~d u=2 td t\]
Use substitution: Let t^2 = m then 2tdt = dm The new integral is; \[\int\sqrt{1-m^2}dm\]

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Other answers:

\[\int\limits \sqrt{1-u^2} \, du\] u=sin(a) \[\sqrt{1-u^2}\text{ = }\sqrt{1-\sin ^2(s)}\text{ = }\cos (s)\text{ and }s=\sin ^{-1}(u)\]
\[\int\limits \cos ^2(s) \, ds\] half angle \[\int\limits \left(\frac{1}{2} \cos (2 s)+\frac{1}{2}\right) \, ds\] separate \[\int\limits \frac{1}{2} \, ds+\frac{1}{2}\int\limits \cos (2 s) \, ds\] solve
you are right but I obtained it from: \[\int\limits \cos^2x \sqrt{sinx} dx\]
@.Sam. don't understand how I get it!
@shivam_bhalla help me dear!
i didn't get \[\int\limits\limits \cos^2x \sqrt{sinx} dx\]
@RedPrince , follow what @.Sam. told
@.Sam. integrate it! I get the result \[\int\limits \sqrt{1-t^4}.2tdt\] from \[∫\cos^2x \sqrt{sinx}dx\] by substituting \[sinx=t^2\]
@shivam_bhalla I also follow. you also?
its a long step to integrate this \[∫\cos^2x \sqrt{sinx}dx\]
I simply remember that \[\int\limits_{}^{}\sqrt{a^2-x^2} = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}{(x/a)}\]
@.Sam. you guide me from where I ask question! see my uestion
@RedPrince , see the above integral I wrote, substitute and simplify
@shivam_bhalla this is good and now the question is solved.....!! woooowwwwWWw. Thanksss!
you could use this too
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@.Sam. , was just writing the same :)
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@.sam yes you are right. this is also helpful.
but that's just the formula without steps
@RedPrince , I suggest you remember these formulas because they take a long time to derive. This is just formulas to quickly solve problems
But , however I suggest you try to derive these formulas on your own first and then remember them
@.Sam. it doesn't matter. I will prove them myself. again thanks!
@shivam_bhalla yes, I will remember these formulas Inshallah.
RedPrince: I /think/ you may have made a mistake in your substitutions. I get this after the substitutions:\[\int\limits \sqrt{1-t^4}.2t^2dt\]
@asnaseer , hawk eye :)
So our new integral should be \[m=t^2\] \[dm = 2tdt\] \[\int\limits_{}^{}\sqrt{1-m^2}\sqrt{m}dm\]
I /believe/ the original integral is not as straightforward as it might look. this is what wolfram thinks the answer should be:^2%28x%29*sqrt%28sin%28x%29%29
@asnaseer yes, you are right but how can I solve it i.e., \[\int\limits \sqrt{m-m^3}\]
see the link I posted - it is not an easy integral (even in this form):^3%29
you need an expert like @JamesJ - he might be able to see a way to getting to the answer.
@asnaseer O.K., just see who help me in solving this question!
@RedPrince , the qustion is mis understood. the first solution is the correct one
@vamgadu have you read ALL the comments above? the original equation to solve was:\[∫\cos^2x \sqrt{sinx}dx\]this was incorrectly transformed to:\[\int\limits \sqrt{1-t^4}.2tdt\]
@FoolForMath help please!
@heena @amistre64 @AccessDenied @satellite73 help plaese if you have idea!
just integrate by parts
(1+t^4)^1/2 2t ..... hmm, seems i confused my integrating and deriving terms again -2
its still by parts but a trig sum does seem plausible
let tan(u) = t^2 sec^2(u) du = 2t dt replace the substitutions\[\int \sqrt{1+tan^2(u)}\ du\]
really need to wake up this morning :) \[\int \sqrt{1+tan^2(u)}\ sec^2(u)du\]
that would have worked out if I hadnt misread the operator; but same concept using the a sin or cos instead of a tangent

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