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

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

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

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

.Sam.
 2 years ago
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

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

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

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

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

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

RedPrince
 2 years ago
Best 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 \[sinx=t^2\]

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

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

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

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

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

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

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

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

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

shivam_bhalla
 2 years ago
Best 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

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

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

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

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

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

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

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

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3I /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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.3see 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

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

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

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

asnaseer
 2 years ago
Best 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:\[\int\limits \sqrt{1t^4}.2tdt\]

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

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

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

amistre64
 2 years ago
Best 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

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

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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1that 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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