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  1. integrate sinx+cosx/sqr root sinxcosx

 So, equation becomes,

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 write denominator : sq root of sinx cosx = 2sinx cosx/2 =1/2* ( 1 - (1- (2sinxcosx)) = 1/2*(1- ( sin2x+Cos2x -2sinxcosx) )

=1/2*(1- ( sinx-Cosx)2 . so expression becomes: integr of  (Sinx + Cos x)/ 1/2*(1- ( sinx-Cosx)2,)

now put Sinx- Cosx = 't' => ( Cosx + Sinx  )dx = dt , =>  (sinx + Cosx) dx = dt

so exprssion becomes: dt/ 1/2*(1- t2) = 2dt/(1-t2) = 2 *1/2 log  !(1+t)/(1-t)! +C

Now put t = Sinx - Cosx in this result & get the answer.

  • -5

write denominator : sq root of sinx cosx = 2sinx cosx/2=1/2* (1 - (1- (2sinxcosx)) = 1/2*(1- (sin2x+Cos2x -2sinxcosx) )

 write sinx*cosx = 2sinx cosx/2=1/2* (1 - (1- (2sinxcosx)) = 1/2*(1- (sin2x+Cos2x -2sinxcosx) )

=1/2*(1- ( sinx-Cosx) 2 .

so expression becomes: integr of (Sinx + Cos x)/ sq root of 1/2*(1- ( sinx-Cosx) 2, )

now put Sinx- Cosx = 't' => (Cosx + Sinx )dx = dt , => (sinx + Cosx) dx = dt

so exprssion becomes: dt/ sq root of  1/2*(1- t2) = (sq rt of 2)dt/sq rt of (1-t2) = (sq rt of 2) *sin -1(t) +C

(In my earlier comment i forgot to considr sq root)

Now put t = Sinx - Cosx in this result & get the answer.

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