Find value of integral:   `int_(pi/4)^(..5pi/12)(sen^4x-cos^4x)/sqrt[sen2x(sen(pi/3+2x)-sen(2x-pi/3))] dx`

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pramodpandey | College Teacher | (Level 3) Valedictorian

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`sqrt(cot(y))-sqrt(tan(y))=(cos(y)-sin(y))/sqrt(sin(y)cos(y))=(sqrt(2)(cos(y)-sin(y)))/sqrt(2sin(y)cos(y))`

`=(sqrt(2)(cos(y)-sin(y)))/sqrt((sin(y)+cos(y))^2-1)`

let  sin(y)+cos(y)=t

dt=(cos(y)-sin(y))dy

`=int(sqrt(2)(cos(y)-sin(y)))/sqrt((sin(y)+cos(y))^2-1)dy=int(sqrt(2)dt)/sqrt(t^2-1)`

`` let t=sec(a)

dt =sec(a)tan(a)da

`int(sqrt(2)dt)/sqrt(t^2-1)=int(sqrt(2)(sec(a)tan(a)da))/tan(a)`

`=sqrt(2)log(sec(a)+tan(a))`

`=sqrt(2)log(t+sqrt(t^2-1))`

`=sqrt(2)log(sin(y)+cos(y)+sqrt(2sin(y)cos(y)))`

`` `int_(pi/2)^((5pi)/6)sqrt(cot(y))/(2(3)^(1/4))dy=`

`sqrt(2){sin^(-1)(sin(y)-cos(y))+`

`log(sin(y)+cos(y)+sqrt(2sin(y)cosy)))}_(pi/2)^((5pi)/6)`

Please put limits and simplify.

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pramodpandey | College Teacher | (Level 3) Valedictorian

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sorry again I slipped

whole right side get divided by`2(3)^(1/4)`  ,then we get correct ans.

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pramodpandey | College Teacher | (Level 3) Valedictorian

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`int_(pi/4)^((5pi)/12)(sin^4x-cos^4x)/(sqrt(sin(2x)(sin(pi/3+2x)-sin(2x-pi/3))))dx`

`` simplify integrand ,we have

`int_(pi/4)^((5pi)/12)sqrt(cot(2x))/((3)^(1/4))dx`

let  2x=y , 2dx=dy

x=pi/4 , y=pi/2

x=(5pi)/12 , y=(5pi)/6

`=int_(pi/2)^((5pi)/6)sqrt(cot(y))/(2(3)^(1/4))dx`

`2sqrt(cot(y))=(sqrt(cot(y))+sqrt(tan(y)))+(sqrt(cot(y))-sqrt(tan(y)))`

`sqrt(cot(y))+sqrt(tan(y))=(cos(y)+sin(y))/sqrt(sin(y)cos(y))`

`=(sqrt(2)(cos(y)+sin(y)))/sqrt(1-(sin(y)-cos(y))^2)`

let `sin(y)-cos(y)=t`

`-dt= (sin(y)+co(y))dy`

`intsqrt(2)(-dt)/sqrt(1-t^2)=sqrt(2)sin^(-1)(t)`

`=sqrt(2)sin^(-1){sin(y)-cos(y)}`

Similarly we can consider otherpart and integrate  complete question.

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