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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;
\[\int\sqrt{1-m^2}dm\]

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

@shivam_bhalla I also follow. you also?

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

@.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

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

or

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

but that's just the formula without steps

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.

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

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

:)

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

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

@FoolForMath help please!

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