Question

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

Answer 1

\[\int\limits \sqrt{1-t^4}.2tdt\]

Answer 2

\[u=t^2~~~and ~~~d u=2 td t\]

Answer 3

Use substitution:
Let t^2 = m
then 2tdt = dm
The new integral is;
\[\int\sqrt{1-m^2}dm\]

Answer 4

\[\int\limits \sqrt{1-u^2} \, du\]
u=sin(a)
\[\sqrt{1-u^2}\text{ = }\sqrt{1-\sin ^2(s)}\text{ = }\cos (s)\text{ and }s=\sin ^{-1}(u)\]

Answer 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

Answer 6

you are right but I obtained it from:
\[\int\limits \cos^2x \sqrt{sinx} dx\]

Answer 7

?

Answer 8

@.Sam. don't understand how I get it!

Answer 9

@shivam_bhalla help me dear!

Answer 10

i didn't get \[\int\limits\limits \cos^2x \sqrt{sinx} dx\]

Answer 11

@RedPrince , follow what @.Sam. told

Answer 12

@.Sam. integrate it!
I get the result \[\int\limits \sqrt{1-t^4}.2tdt\]
from \[∫\cos^2x \sqrt{sinx}dx\]
by substituting \[sinx=t^2\]

Answer 13

@shivam_bhalla I also follow. you also?

Answer 14

its a long step to integrate this
\[∫\cos^2x \sqrt{sinx}dx\]

Answer 15

I simply remember that
\[\int\limits_{}^{}\sqrt{a^2-x^2} = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}{(x/a)}\]

Answer 16

@.Sam. you guide me from where I ask question! see my uestion

Answer 17

@RedPrince , see the above integral I wrote, substitute and simplify

Answer 18

@shivam_bhalla this is good and now the question is solved.....!!
woooowwwwWWw. Thanksss!

Answer 19

you could use this too

Answer 20

@.Sam. , was just writing the same :)

Answer 21

or

Answer 22

@.sam yes you are right. this is also helpful.

Answer 23

but that's just the formula without steps

Answer 24

@RedPrince , I suggest you remember these formulas because they take a long time to derive. This is just formulas to quickly solve problems

Answer 25

But , however I suggest you try to derive these formulas on your own first and then remember them

Answer 26

@.Sam. it doesn't matter. I will prove them myself.
again thanks!

Answer 27

@shivam_bhalla yes, I will remember these formulas Inshallah.

Answer 28

RedPrince: I /think/ you may have made a mistake in your substitutions.
I get this after the substitutions:\[\int\limits \sqrt{1-t^4}.2t^2dt\]

Answer 29

\[\sin(x)=t^2\]therefore:\[\cos(x)dx=2tdt\]

Answer 30

\[\cos^2(x)=1-\sin^2(x)=1-t^4\]

Answer 31

therefore:\[\cos^2(x)\sqrt{sin(x)}dx=\cos^2(x)t*\frac{2tdt}{cos(x)}=\cos(x)2t^2dt\]\[\qquad=\sqrt{1-t^4}2t^2dt\]

Answer 32

@asnaseer , hawk eye :)

Answer 33

:)

Answer 34

So our new integral should be
\[m=t^2\]
\[dm = 2tdt\]
\[\int\limits_{}^{}\sqrt{1-m^2}\sqrt{m}dm\]

Answer 35

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

Answer 36

@asnaseer yes, you are right but how can I solve it i.e.,
\[\int\limits \sqrt{m-m^3}\]

Answer 37

see the link I posted - it is not an easy integral (even in this form): http://www.wolframalpha.com/input/?i=integrate+sqrt%28x-x^3%29

Answer 38

you need an expert like @JamesJ - he might be able to see a way to getting to the answer.

Answer 39

@asnaseer O.K., just see who help me in solving this question!

Answer 40

@RedPrince , the qustion is mis understood. the first solution is the correct one

Answer 41

@vamgadu have you read ALL the comments above?
the original equation to solve was:\[∫\cos^2x \sqrt{sinx}dx\]this was incorrectly transformed to:\[\int\limits \sqrt{1-t^4}.2tdt\]

Answer 42

@FoolForMath help please!

Answer 43

@heena @amistre64 @AccessDenied @satellite73 help plaese if you have idea!

Answer 44

just integrate by parts

Answer 45

(1+t^4)^1/2
2t ..... hmm, seems i confused my integrating and deriving terms again
-2

Answer 46

its still by parts but a trig sum does seem plausible

Answer 47

let tan(u) = t^2
sec^2(u) du = 2t dt

replace the substitutions\[\int \sqrt{1+tan^2(u)}\ du\]

Answer 48

really need to wake up this morning :)

\[\int \sqrt{1+tan^2(u)}\ sec^2(u)du\]

Answer 49

that would have worked out if I hadnt misread the operator; but same concept using the a sin or cos instead of a tangent