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# (a) int of -arc tan root x is(b) int of-cos^4 x is Grade:12

## 1 Answers

11 years ago

(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

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