Integrate the following:- ∫(sin^6 x+cos^6 x)/sin^2 x.cos^2 x dx

Question

Arun · Answer

First note that sin 6 ( x ) + cos 6 ( x ) is a sum of two cubes ( sin 2 ( x ) ) 3 + ( cos 2 ( x ) ) 3 so it can be factored using a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) and the Pythagorean identity to get sin 6 ( x ) + cos 6 ( x ) = ( sin 2 ( x ) ) 3 + ( cos 2 ( x ) ) 3 = ( sin 2 ( x ) + cos 2 ( x ) ) ( sin 4 ( x ) − sin 2 ( x ) cos 2 ( x ) + cos 4 ( x ) ) = sin 4 ( x ) − sin 2 ( x ) cos 2 ( x ) + cos 4 ( x ) . Therefore, sin 6 x + cos 6 x sin 2 ( x ) cos 2 ( x ) = sin 4 ( x ) − sin 2 ( x ) cos 2 ( x ) + cos 4 ( x ) sin 2 ( x ) cos 2 ( x ) = tan 2 ( x ) − 1 + cot 2 ( x ) But tan 2 ( x ) = sec 2 ( x ) − 1 and cot 2 ( x ) = csc 2 ( x ) − 1 . Hence, sin 6 x + cos 6 x sin 2 ( x ) cos 2 ( x ) = sec 2 ( x ) + csc 2 ( x ) − 3 and thus ∫ sin 6 ( x ) + cos 6 ( x ) sin 2 ( x ) cos 2 ( x ) d x = ∫ sec 2 ( x ) + csc 2 ( x ) − 3 d x = tan ( x ) − cot ( x ) − 3 x + C

tuhar · Answer

=sin 6 ( x )/sin 2 x.cos 2 x + cos 6 ( x ) /sin 2 x.cos 2 x =sin 4 x/cos 2 x+cos 4 x/sin 2 x =(1-cos 2 x) 2 /cos 2 x + (1-sin 2 x) 2/ sin 2 x =(1-2cos 2 x+cos 4 x)/cos 2 x + (1-2sin 2 x+sin 4 x)/sin 2 x =sec 2 x -2 + cos 2 x + cosec 2 x -2 + sin 2 x = sec 2 x + cosec 2 x -4 + 1 = sec 2 x + cosec 2 x -3 so, by integration we get, =tan x – cot x -3x + C

Rishi Sharma · Answer

Dear Student,
Please find below the solution to your problem.

first resolve , sin^6x + cos^6x
= (sin²x)³ + (cos²x)³
= (sin²x + cos²x)(sin⁴x + cos⁴x - sin²x.cos²x)
=(sin²x + cos²x) {(sin²x + cos²x)² - 3sin²x.cos²x}
but we know, sin²x + cos²x = 1
so, sin^6x + cos^6x = (1 - 3sin²x.cos²x)
or, you can write cos^6x + cos^6x = sin²x + cos²x - 3sin²x.cos²x [ as we know, sin²x + cos²x = 1 ]

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Integrate the following:- ∫(sin^6 x+cos^6 x)/sin^2 x.cos^2 x dx

Integrate the following:-
∫(sin^6 x+cos^6 x)/sin^2 x.cos^2 x dx

3 Answers

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Integrate the following:- ∫(sin^6 x+cos^6 x)/sin^2 x.cos^2 x dx

Integrate the following:-∫(sin^6 x+cos^6 x)/sin^2 x.cos^2 x dx

3 Answers

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Integrate the following:-
∫(sin^6 x+cos^6 x)/sin^2 x.cos^2 x dx