IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ans

_{[ ]}

^{[ ]} sin2xdx/sin⁶x+cos^6x

cooldude

use thatsin^6 x+cos^6 x=3/4(cos^2 2x)+1/4 and put cos2x=t.

aditya_arora04

Hi,

Look,

I am not using the integration sign. Assume it is present there.

sin2x / ( sin⁶x + cos⁶x )

As, a³ + b³f = (a+b)(a² +b² - ab )

sin2x / [ (sin²x + cos²x )(sin⁴x + cos⁴x - sin²xcos²x ) ]

As, sin²x + cos²x = 1

sin2x / [ (sin²x + cos²x )² - 2sin²xcos²x - sin²xcos²x ]

sin2x / [ 1 - 3sin²xcos²x ]

Divide num and denom by cos⁴x. And yes, open sin2x.

2Tanx*sec²x / [ sec⁴x - 3Tan²x ]

Write (Tan²x + 1)² in place of sec⁴x. Open it.

Now, put t = tanx [ substitution and find the answer, ans !!! ].

Hope you have got it !!!

aditya_arora04

And one more thing which i forgot to tell :

After substitution, you will get :

2t dt / [ t⁴ - t² + 1 ]

Here, do another substituion by putting k = t².

We get :

dk / [ k² - k + 1 ]

Then make a perfect square and use formulae for special integration.

Then you get the answer !!!!

edison

well done aditya.

dhashrath

well not using integration Sign
Simplify Usin addition of cubes a³+b³.............................
then you will get
sin 2x/(1-3sin²xcos²x)
then substtiute Cos²x as (1-Sin²x)
and put sin²x as 't' and sin2x dx as dt

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