int. calculus \[\int\limits_{}^{}\sin ^{2}x \cos ^{2}x \tan ^{2} x dx\]

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int. calculus \[\int\limits_{}^{}\sin ^{2}x \cos ^{2}x \tan ^{2} x dx\]

Mathematics
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Write them as sin squared x = (1/2)(1 - cos 2x) cos squared x = (1/2)(1 + cos 2x) maybe..that'll be easier for you to integrate.
can i do that, even if sin or cos is not raised two?
doesn't that simplify to sin^4 x when tan is written as sin/cos ?

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Write tan^2 x = sin^2 x / cos^2 x
yea
i did that before
\[\large{\int { sin^4x dx}}\] Can you do this?
then i lost any of ideas after that
OK no problem, where are you lost?
can u integrate sin^2 x ?
after i got the \[\int\limits_{?}^{?}\sin ^{4}xdx\]
Well if integration of sin x dx= -cosx + C then that of sin^4 x dx = ?
-cos^4x + C ??
write sin^4 x = (sin^2 x)(1-cos^2 x)
no, its not that direct
then distribute?
= (sin^2 x)(1-cos^2 x) = sin^2 x - sin^2x cos^2 x yes
= sin^2 x - sin^2x cos^2 x = sin^2 x - (1/4) (sin 2x)^2
how did you get (1/4)(sin 2x)^2?
sin 2x = 2 sin x cos x to get sin^2x cos^2 x , i squared the above .....
got that ? ^ and can u integrate sin^2 x - (1/4) (sin 2x)^2
can you show me the whole process of that sin 2x = 2 sin x cos x
u want me to prove 2sin x cos x = sin 2x or u wanted this : sin 2x = 2 sin x cos x (sin 2x)^2 = (2 sin x cos x)^2 = 4 sin^2x cos^2 x so sin^2 x cos^2 x = (1/4) (sin 2x)^2
so if u can integrate sin^2 x you can also integrate sin^2 x - (1/4) (sin 2x)^2

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