Chapter 5, Section 9   Inverse Trigonometric Functions: Integration

Integrals involving Inverse Trig Functions

The derivatives of the six inverse trigonometric functions fall into 3 pairs.  

In each pair, the derivative of one function is the negative of the other.  For example,   

(d/dx)[arcsin x] =   / (1 - x)^()   

 and

dy/dx [arccos u] = - / (1 - x)^()

 

When listing the antiderivative that corresponds to each of the inverse trigonometric functions, you need to use only one member of each pair.  It is conventional to use arcsin x as the antiderivative of (1 / (sqrt(1 - x))), rather than arcos x.  The next theorem gives one antiderivative formula for each of the three pairs.  

Theorem 5.19:   Integrals involving Inverse Trigonometric Functions.

Let u be a differentiable function of x, and let a > 0 

    + C    **Remember that arcsin is equal to sin - 1 .
   + C     ***Remember that arctan is equal to tan - 1 .
   + C    ****Remember that arcsec is equal to sec - 1 .

EXAMPLE 1    Integration with Inverse Trig Functions

(a)  ∫ (dx / ( sqrt(4 - x)) 

since u =  x   and   a = 2  then  

∫ (dx / ( sqrt(4 - x) )  = arcsin (x/2) + C 


(b)  ∫ (dx /  (2 + 9) ) ↔ (1/3) ∫ (3dx)/ ((sqrt2) + (3x) )   

since u =  3x  and   a = sqrt2  then 

 ∫ (dx /  (2 + 9) )   = 1/(3*sqrt2) arcstan (3x / sqrt 2)   + C 


(c)  ∫ (dx /( x*  sqrt(4x - 9) )) ↔  ∫ (2dx)/ (2x* sqrt((2x) - 3 )  

 since u =  2x  and   a = 3   then 

 ∫ (dx /( x*  sqrt(4x - 9) )) = (1/3) arcsec ( |2x| / 3 ) + C

 


EXAMPLE 2    Integration by Substitution

find ∫ (dx)/(sqrt(ex - 1))

Solution:  As it stands, this integral doesn't fit any of the three inverse trigonometric formulas.  Using the substitution u = ex , however, produces the following.

u =  ex    →   du = ex dx    →   dx = (du / ex) = (du / u)

With this substitution, you can integrate as follows.

∫ (dx)/(sqrt(ex - 1)) = ∫ (dx)/(sqrt( (ex) - 1))  

       = ∫ (du/u)/(sqrt(u - 1))   ...Substitution

      = ∫ (du/u)/(u*(sqrt(u - 1)) )  ...Rewrite to fit Arcsecant Rule

      = arcsec (|u|/1) + C  ...Apply Arcsecant Rule

       = arcsec  ex   + C   ...Back Substitution

 

EXAMPLE 3    Rewriting as the Sum of Two Quotients

find ∫ (x + 2)/(sqrt(4 - x ))dx

Solution   This integral does not appear to fit any of the basic integration formulas.  By splitting the integrand into two parts, however, you can see the first part can be found with the Power Rule and the second part yields an inverse sine function.

 ∫ (x + 2)/(sqrt(4 - x ))dx =  ∫ (x)/(sqrt(4 - x ))dx +  ∫ ( 2)/(sqrt(4 - x ))dx

          = (-1/2)* ∫ (4 - x )^(-1/2)(-2x) dx + 2* ∫ ( 1)/(sqrt(4 - x ))dx

          = (-1/2)* [(4 - x )^(-1/2)/ (1/2)]* 2 arcsin( x/2 ) + C

          = - (sqrt(4 - x ) + 2 arcsin( x/2 ) + C


Review of Basic Integration Rules   

See this link for     Common Integrals

The chart below represents some of the formulas that is on that link.

1.
$\displaystyle \int adx=ax$
2.
$\displaystyle \int af(x)dx=a \displaystyle \int f(x)dx$
3.
$\displaystyle \int \left( u \pm v \pm w \pm \cdots \right) dx =
\displaystyle \int udx \pm \displaystyle \int vdx \pm \displaystyle
\int wdx \pm \cdots $
4.
$\displaystyle \int udv = uv - \displaystyle \int vdu$
5.
$\displaystyle \int f(ax)dx = \displaystyle \frac{1}{a} \displaystyle
\int f(u)du$
6.
$\displaystyle \int F\{f(x)\}dx = \displaystyle \int F(u) \displaystyle
\frac{dx}{du}du = \displaystyle \int \displaystyle
\frac{F(u)}{f'(x)}du$
7.
$\displaystyle \int u^{n}du = \displaystyle \frac{u^{n+1}}{n+1}, n \neq -1$
8.
$\begin{array}{lcl}
\displaystyle \int\displaystyle \frac{du}{u} & = & \ln u \mb...
...or} \ln (-u) \mbox{ if} u<0 \\
& = & \ln \left\vert u \right\vert
\end{array}$
9.
$\displaystyle \int e^{u}du=e^{u}$
10.
$\displaystyle \int a^{u}du = \int e^{u \ln a}du = \displaystyle \frac{e^{u \ln a}}{\ln a} = \displaystyle \frac{a^{u}}{\ln a} , a >0, a \neq 1$
11.
$\displaystyle \int \sin u du = -\cos u$
12.
$\displaystyle \int \cos u du = \sin u$
13.
$\displaystyle \int \tan u du = \ln \sec u = -\ln \cos u$
14.
$\displaystyle \int \cot u du = \ln \sin u$
15.
$\displaystyle \int \sec u du = \ln (\sec u + \tan u) = \ln \tan \left( \displaystyle \frac{u}{2} + \displaystyle \frac{\pi}{4} \right)$
16.
$\displaystyle \int \csc u du = \ln (\csc u - \cot u) = \ln \tan \displaystyle \frac{u}{2}$
17.
$\displaystyle \int \sec ^{2} u du = \tan u$
18.
$\displaystyle \int \csc ^{2} u du = -\cot u$
19.
$\displaystyle \int \tan ^{2} u du = \tan u - u$
20.
$\displaystyle \int \cot ^{2} u du = -\cot u - u $
21.
$\displaystyle \int \sin ^{2} u du = \displaystyle \frac{u}{2} - \displaystyle \frac{\sin 2u}{4} = \displaystyle \frac{1}{2} (u-\sin u \cos u)$
22.
$\displaystyle \int \cos ^{2} u du = \displaystyle \frac{u}{2} + \displaystyle \frac{\sin 2u}{4} = \displaystyle \frac{1}{2} (u+\sin u \cos u)$
23.
$\displaystyle \int \sec u \tan u du = \sec u$
24.
$\displaystyle \int \csc u \cot u du = -\csc u $
25.
$\displaystyle \int\displaystyle \frac{du}{u^{2}+a^{2}} = \displaystyle \frac{1}{a} \tan^{-1} \displaystyle \frac{u}{a}$
26.
$\displaystyle \int\displaystyle \frac{du}{\sqrt{a^{2}-u^{2}}} = \sin ^{-1} \displaystyle \frac{u}{a}$
27.
$\displaystyle \int\displaystyle \frac{du}{\sqrt{u^{2}-a^{2}}} = \ln \left( u + \displaystyle\sqrt{u^{2} - a^{2}} \right)$
28.
$\displaystyle \int\displaystyle \frac{du}{u \sqrt{u^{2}-a^{2}}} = \displaystyle \frac{1}{a} \sec ^{-1} \left\vert \displaystyle \frac{u}{a} \right\vert$
29.
$\displaystyle \int\displaystyle \frac{du}{u \sqrt{u^{2}+a^{2}}}=-\displaystyle \frac{1}{a} \ln \left( \displaystyle \frac{a+\sqrt{u^{2}+a^{2}}}{u} \right)$
30.
$\displaystyle \int\displaystyle \frac{du}{u \sqrt{a^{2}-u^{2}}}=-\displaystyle \frac{1}{a} \ln \left( \displaystyle \frac{a+\sqrt{a^{2}-u^{2}}}{u} \right)$
 

For homework examples, see Assignment #1