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mustafa bandook wala Grade: 12
        

(a) int of -arc tan root x is


(b) int of-cos^4 x is


 

8 years ago

Answers : (1)

Ramesh V
70 Points
										

(a)  int of: tan-1x1/2.dx


take tan-1x1/2 = t


dx/(1+x).2x1/2 = dt


so tan t = x1/2  or x = tan2x


= int of: 2t.tan t(1+tan2t).dt


= int of: 2t.tan t.sec2t.dt


integrating by parts gives :


= 2 [ t.tant2t/2 - int( (tan2t) /2 .dt) ]


= t.tan2t - int( (sec2t) -1).dt


= t.tan2t - tan t + t +C  


=x.tan-1x1/2 - x1/2 + tan-1x1/2 + C


=(x+1)tan-1x1/2 - x1/2  + C :where C is constant


 


 


(b)  int of: cos^4x.dx


Note that:   1 + cos 2A = 2cos2A


so  1 + cos 2x = 2cos2x


   =int of :  ((1 + cos 2x) / 2 )2


   = int of :  1/4*(1 + 2cos 2x+ cos22x).dx


   = int of :  [ 1/4*(1 + 2cos 2x).dx +1/8*(1 + cos 4x).dx ]


   =x/4 + sin 2x /4 +x/8 +sin 4x /32 +C where C is constant


 


   =3x/8 + (sin 2x)/4 + (sin 4x)/32 +C         :where C is constant

8 years ago
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