Page 38 - e-book Calculcs I

1
2
= 3 2 ln 3 (2 ) + 10 3 + ln (30 )

2
2
= 2 (3 2 ln 3) + 10 + 30 ln

2
−1
2
5.3 ( ) = sin + tan (√ + 1) (4 คะแนน)
1+cos
2
2
(1+cos ) sin − sin (1+cos ) 1 1
′
2
( ) = + ( + 1)2
2
(1+cos ) 2 1+(√ +1)
2

2
(1+cos )(2 sin ) sin − sin sin 1 1 − 1
2
2
= + ( + 1) 2 ( + 1)
2
(1+cos ) 2 +2 2
3
= (1+cos )(2 sin ) cos − sin + 2
2
2
(1+cos ) 2 2( +2)√ +1

5.4 ( ) = (cos ) sin 2 (4 คะแนน)
ให้ = (cos ) sin 2
∴ ln = ln(cos ) sin 2
2
= (sin )(ln cos )
1
2
= sin 2 ln cos + ln cos sin

1

= sin 2 cos cos + ln(cos ) cos 2 2

2
2
= (sin )(− sin ) + ln(cos )(cos )(2 )
cos
2
(sin )(− sin ) 2
2
∴ = [ + ln(cos )(cos )(2 )] (cos ) sin
cos

6. ก าหนดให้ cos + sin = 4 + 1 จงหา (5 คะแนน)

( cos + sin ) = (4 + 1)

cos + cos + sin + sin = 4 + 4

− sin + cos + cos + sin = 4 + 4

− sin + sin − 4 = 4 −cos − cos

(− sin + sin − 4 | = 4 −cos − cos

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