342 DIFFERENTIAL CALCULUS
- (a) sech−^1 (x−1) (b) tanh−^1 (tanhx)
[
(a)
− 1
(x−1)
√
[x(2−x)]
(b) 1
]
- (a) cosh−^1
(
t
t− 1
)
(b) coth−^1 (cosx)
[
(a)
− 1
(t−1)
√
(2t−1)
(b)−cosecx
]
- (a)θsinh−^1 θ (b)
√
xcosh−^1 x
⎡
⎢
⎢
⎢
⎣
(a)
θ
√
(θ^2 +1)
+sinh−^1 θ
(b)
√
x
√
(x^2 −1)
+
cosh−^1 x
2
√
x
⎤
⎥
⎥
⎥
⎦
- (a)
2 sec h−^1
√
t
t^2
(b)
tan h−^1 x
(1−x^2 )
⎡ ⎢ ⎢ ⎢ ⎢ ⎣
(a)
− 1
t^3
{
1
√
(1−t)
+4 sech−^1
√
t
}
(b)
1 + 2 xtanh−^1 x
(1−x^2 )^2
⎤ ⎥ ⎥ ⎥ ⎥ ⎦
- Show that
d
dx
[xcosh−^1 (coshx)]= 2 x
In Problems 13 to 15, determine the given
integrals
- (a)
∫
1
√
(x^2 +9)
dx
(b)
∫
3
√
(4x^2 +25)
dx
[
(a) sinh−^1
x
3
+c(b)
3
2
sinh−^1
2 x
5
+c
]
- (a)
∫
1
√
(x^2 −16)
dx
(b)
∫
1
√
(t^2 −5)
dt
[
(a) cosh−^1
x
4
+c(b) cosh−^1
t
√
5
+c
]
- (a)
∫
dθ
√
(36+θ^2 )
(b)
∫
3
(16− 2 x^2 )
dx
⎡
⎢
⎢
⎣
(a)
1
6
tan−^1
θ
6
+c
(b)
3
2
√
8
tanh−^1
x
√
8
+c
⎤
⎥
⎥
⎦