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RedPrince Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits \sqrt{1t^4}.2tdt\]
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.5
\[u=t^2~~~and ~~~d u=2 td t\]
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
Use substitution: Let t^2 = m then 2tdt = dm The new integral is; \[\int\sqrt{1m^2}dm\]
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.5
\[\int\limits \sqrt{1u^2} \, du\] u=sin(a) \[\sqrt{1u^2}\text{ = }\sqrt{1\sin ^2(s)}\text{ = }\cos (s)\text{ and }s=\sin ^{1}(u)\]
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.5
\[\int\limits \cos ^2(s) \, ds\] 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
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
you are right but I obtained it from: \[\int\limits \cos^2x \sqrt{sinx} dx\]
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@.Sam. don't understand how I get it!
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@shivam_bhalla help me dear!
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.5
i didn't get \[\int\limits\limits \cos^2x \sqrt{sinx} dx\]
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@RedPrince , follow what @.Sam. told
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@.Sam. integrate it! I get the result \[\int\limits \sqrt{1t^4}.2tdt\] from \[∫\cos^2x \sqrt{sinx}dx\] by substituting \[sinx=t^2\]
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@shivam_bhalla I also follow. you also?
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.5
its a long step to integrate this \[∫\cos^2x \sqrt{sinx}dx\]
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
I simply remember that \[\int\limits_{}^{}\sqrt{a^2x^2} = \frac{x}{2}\sqrt{a^2x^2}+\frac{a^2}{2}\sin^{1}{(x/a)}\]
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@.Sam. you guide me from where I ask question! see my uestion
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@RedPrince , see the above integral I wrote, substitute and simplify
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@shivam_bhalla this is good and now the question is solved.....!! woooowwwwWWw. Thanksss!
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.5
you could use this too
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@.Sam. , was just writing the same :)
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@.sam yes you are right. this is also helpful.
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.5
but that's just the formula without steps
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@RedPrince , I suggest you remember these formulas because they take a long time to derive. This is just formulas to quickly solve problems
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
But , however I suggest you try to derive these formulas on your own first and then remember them
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@.Sam. it doesn't matter. I will prove them myself. again thanks!
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@shivam_bhalla yes, I will remember these formulas Inshallah.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
RedPrince: I /think/ you may have made a mistake in your substitutions. I get this after the substitutions:\[\int\limits \sqrt{1t^4}.2t^2dt\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
\[\sin(x)=t^2\]therefore:\[\cos(x)dx=2tdt\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
\[\cos^2(x)=1\sin^2(x)=1t^4\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
therefore:\[\cos^2(x)\sqrt{sin(x)}dx=\cos^2(x)t*\frac{2tdt}{cos(x)}=\cos(x)2t^2dt\]\[\qquad=\sqrt{1t^4}2t^2dt\]
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@asnaseer , hawk eye :)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
So our new integral should be \[m=t^2\] \[dm = 2tdt\] \[\int\limits_{}^{}\sqrt{1m^2}\sqrt{m}dm\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
I /believe/ the original integral is not as straightforward as it might look. this is what wolfram thinks the answer should be: http://www.wolframalpha.com/input/?i=integrate+cos^2%28x%29*sqrt%28sin%28x%29%29
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@asnaseer yes, you are right but how can I solve it i.e., \[\int\limits \sqrt{mm^3}\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
see the link I posted  it is not an easy integral (even in this form): http://www.wolframalpha.com/input/?i=integrate+sqrt%28xx^3%29
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
you need an expert like @JamesJ  he might be able to see a way to getting to the answer.
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@asnaseer O.K., just see who help me in solving this question!
 2 years ago

vamgadu Group TitleBest ResponseYou've already chosen the best response.0
@RedPrince , the qustion is mis understood. the first solution is the correct one
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.3
@vamgadu have you read ALL the comments above? the original equation to solve was:\[∫\cos^2x \sqrt{sinx}dx\]this was incorrectly transformed to:\[\int\limits \sqrt{1t^4}.2tdt\]
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@FoolForMath help please!
 2 years ago

RedPrince Group TitleBest ResponseYou've already chosen the best response.0
@heena @amistre64 @AccessDenied @satellite73 help plaese if you have idea!
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
just integrate by parts
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
(1+t^4)^1/2 2t ..... hmm, seems i confused my integrating and deriving terms again 2
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
its still by parts but a trig sum does seem plausible
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
let tan(u) = t^2 sec^2(u) du = 2t dt replace the substitutions\[\int \sqrt{1+tan^2(u)}\ du\]
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
really need to wake up this morning :) \[\int \sqrt{1+tan^2(u)}\ sec^2(u)du\]
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
that would have worked out if I hadnt misread the operator; but same concept using the a sin or cos instead of a tangent
 2 years ago
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