Examples from Section 6.9
Derivatives of Inverse Trignonmetric Functions; Integrals

p. 518

Finding Derivatives

1.

(∂_x) ArcCos[(x^2)]

3.

(∂_t) ArcSin[(Sqrt[2]) t]

5.

(∂_s) ArcSec[2  s + 1]

Expand[%]

7.

(∂_x) ArcCsc[(x^2) + 1]

Expand[%]

13.

(∂_x) Log[ ArcTan[x]]

21.

(∂_x)(x ArcSin[x]  +  (Sqrt[1  -  (x^2)]))

Evaluating Integrals

23.

∫(1/(Sqrt[9 -  (x^2)])) \[DifferentialD]x

25.

∫(1/(17  +  (x^2))) \[DifferentialD]x

27.

∫(1/(x (Sqrt[ 25  (x^2) -  2]))) \[DifferentialD]x

43.

Substitute u  =  (e^x).

∫((E^x)/(1  +  (E^(2  x)))) \[DifferentialD]x

(∫_0^(Log[(Sqrt[3])]))((E^x)/(1  +  (E^(2  x)))) 
        \[DifferentialD]x

45.

Substitute u  =  (y^2).

∫(y/(Sqrt[1  -  (y^4)])) \[DifferentialD]y

53.

Complete the square: (x^2) -  2  x  +  2  =  (((x - 1))^2) +  1.

∫(8/((x^2) -  2  x  +  2)) \[DifferentialD]x

(∫_1^2) (8/((x^2) -  2  x  +  2)) 
        \[DifferentialD]x

Theory and Examples

88.

What is special about the functions

\(f[x_]\  := \ ArcSin[1\/\@\(x\^2 + \ 1\)]; 
g[x_]\  := \ ArcTan[1\/x]; 
    \)

Plot[f[x],  {x, 0, 2}]

Plot[g[x],  {x, 0, 2}]

The functions are equal for x  ≥  0. To see this, draw a right triangle with sides 1, x, and hypothenuse (Sqrt[(x^2) +  1]). If α is the angle opposite to the side 1, then sin (α)  =  (1/(Sqrt[(x^2) +  1]))and tan (α)  =  (1/x). Therefore, f (x)  =  
        (sin^(-1)) ((1/(Sqrt[(x^2) +  1])))  =  
          α  =  
            (tan^(-1)) ((1/x))  =  g (x).
If x  <  0, then f (x)  =  
        f (-x)  =  g (-x)  =  -g (x).

Plot[{f[x],  g[x]},  {x, -2, 0},  
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[0, 1, 0]}]