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RedPrinceBest ResponseYou've already chosen the best response.0
\[\int\limits \sqrt{1t^4}.2tdt\]
 one year ago

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

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

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

.Sam.Best 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
 one year ago

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

RedPrinceBest ResponseYou've already chosen the best response.0
@.Sam. don't understand how I get it!
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@shivam_bhalla help me dear!
 one year ago

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

shivam_bhallaBest ResponseYou've already chosen the best response.1
@RedPrince , follow what @.Sam. told
 one year ago

RedPrinceBest 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 \[sinx=t^2\]
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@shivam_bhalla I also follow. you also?
 one year ago

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

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

RedPrinceBest ResponseYou've already chosen the best response.0
@.Sam. you guide me from where I ask question! see my uestion
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.1
@RedPrince , see the above integral I wrote, substitute and simplify
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@shivam_bhalla this is good and now the question is solved.....!! woooowwwwWWw. Thanksss!
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.1
@.Sam. , was just writing the same :)
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@.sam yes you are right. this is also helpful.
 one year ago

.Sam.Best ResponseYou've already chosen the best response.5
but that's just the formula without steps
 one year ago

shivam_bhallaBest 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
 one year ago

shivam_bhallaBest 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
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@.Sam. it doesn't matter. I will prove them myself. again thanks!
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@shivam_bhalla yes, I will remember these formulas Inshallah.
 one year ago

asnaseerBest 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:\[\int\limits \sqrt{1t^4}.2t^2dt\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
\[\sin(x)=t^2\]therefore:\[\cos(x)dx=2tdt\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
\[\cos^2(x)=1\sin^2(x)=1t^4\]
 one year ago

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

shivam_bhallaBest ResponseYou've already chosen the best response.1
@asnaseer , hawk eye :)
 one year ago

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

asnaseerBest 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
 one year ago

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

asnaseerBest 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
 one year ago

asnaseerBest 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.
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@asnaseer O.K., just see who help me in solving this question!
 one year ago

vamgaduBest ResponseYou've already chosen the best response.0
@RedPrince , the qustion is mis understood. the first solution is the correct one
 one year ago

asnaseerBest 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:\[\int\limits \sqrt{1t^4}.2tdt\]
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@FoolForMath help please!
 one year ago

RedPrinceBest ResponseYou've already chosen the best response.0
@heena @amistre64 @AccessDenied @satellite73 help plaese if you have idea!
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
just integrate by parts
 one year ago

amistre64Best 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
its still by parts but a trig sum does seem plausible
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
let tan(u) = t^2 sec^2(u) du = 2t dt replace the substitutions\[\int \sqrt{1+tan^2(u)}\ du\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
really need to wake up this morning :) \[\int \sqrt{1+tan^2(u)}\ sec^2(u)du\]
 one year ago

amistre64Best 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
 one year ago
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