Question

Integration Q

Answer 1

\[\int\limits_{}^{}\frac{ \sin^2xcos^2x dx}{ (\sin^3x +\cos^3x)^2 }\]

Answer 2

@ganeshie8 @samigupta8 @imqwerty @freckles

Answer 3

i started in one method..but found it to be lengthy..so i need a short method if possible!

Answer 4

Did u try it by opening up applying the formula a^3+b^3 =(a+b)(a^2+b^2-ab)

Answer 5

yup! i did the same thing!

Answer 6

then i converted everything as sin2x...

Answer 7

lol same approach here too

Answer 8

Priyar wud u tell what r u ending up vid dis method

Answer 9

ya wait..

Answer 10

if you expand it using (a+b)^2 then convert the numerator into sin2x you will easily get it

Answer 11

\[\int\limits_{}^{} \frac{ \sin^2 2x dx}{ (1+\sin2x)(4+\sin^2 2x - 4 \sin2x) }\]

Answer 12

@rvc ?

Answer 13

how to proceed?

Answer 14

wait i messed

Answer 15

oh..

Answer 16

wait i found a simpler way..

Answer 17

\(\large \color{black}{\begin{align}
& \int \dfrac{\sin^{2}x \times \cos^{2}x}{(\sin^{3}x + \cos^{3}x)^{2}} dx \hspace{.33em}\\~\\
& =\int \dfrac{\tan^{2}x \times \sec^{2}x}{(1 + \tan^{3}x)^{2}} dx \hspace{.33em}\\~\\
& \normalsize \text{sub }\ \tan x=u \hspace{.33em}\\~\\
\end{align}}\)

Answer 18

thats exactly what i found!!

Answer 19

did the same substitution just now!

Answer 20

then again 1+u^3 = t

Answer 21

anyway thanks all!

Answer 22

thank you :)
you are helping me revise :)

Answer 23

oh welcome!

Answer 24

lol me too haha nice :)

Answer 25

lol qwerty

Answer 26

so..one more question?

Answer 27

alright :)

Answer 28

thanks..

Answer 29

yay
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