IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ankeet_gala

hello ,
this is ankeet p gala frm vadodara
i would like to seek your help on this two problems

1. Integral of ( sin 7x/ sin x)

2. Integral of (1+x^4/ 1+x^6)

thank you.
apg
vadodara
gujarat

vinu

Hi ankeet_gala,

1). write sin 7x = sin (3x+4x) ;

=sin 3xcos 4x +sin 4xcos 3x

{ sin 4x=4sin xcos xcos 2x ; sin 3x=sinx(3 -4sin²x) = sinx(2cos2x+1) }

Therefore,sin 7x/sinx =4cosxcos2xcos3x + cos4x(2cos2x+1) ;

Now,4cosxcos2xcos3x = 1+cos2x+cos4x+cos6x ;

2cos2xcos4x = cos6x + cos2x ;

Hence,

sin 7x / sin x =

1+ 2cos2x +cos4x +2cos6x ;

which u can integrate easily .

magiclko

well vinu is absolutely correct, as always

....and the scond one can be done by partial fraction method...

deeksha

HI ANKEET

DO IT AS FOLLOWS

THE FACTORS OF x^6+1 = (x^2+1)(x^4-x^2+1)

so add and subtract x^2 IN THE NR

I=

{x^4-x^2+1+x^2} /{x^6+1}

I=

1/{x^2+1}dx +1/3

{3x^2}dx/{[x^3]^2+1}

I=TAN[INV] X +{1/3}TAN[INV}X^3

WHERE ^MEANS POWER..................................

deepak_agarwal

students have done both the question for me...erfect...cheers

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