Substitution rules.

(A.1).
$ \displaystyle\int y \frac{dt}{dx} dx = \int y dt$

  (a) $\displaystyle \int\frac{\sin\sqrt{x}}{\sqrt{x}} dx = 2\int \sin t dt \hspace{...
...ox{(A.1) $t = \sqrt{x}$,\: $\frac{dt}{dx} = \frac{1}{2\sqrt{x}}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = -2 \cos t + C = -2 \cos\sqrt{x} + C$    

  (b) $\displaystyle \int \frac{\cos(\ln x)}{x} dx = \int \cos t dt \hspace{0.8in}\fbox{(A.1) $t = \ln x$,\: $\frac{dt}{dx} = \frac{1}{x}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \sin t + C = \sin(\ln x) + C$    

  (c) $\displaystyle \int \frac{\sin x}{1 + \sin x} dx = \int \frac{\sin x (1 - \sin x)}{1 - \sin^2 x} dx$    
  $\displaystyle \hspace{0.4in} = \int\frac{\sin x}{\cos^2 x} dx - \int\frac{\sin...
...pace{0.4in}\fbox{(A.1) $t = \cos x$,\: $\frac{dt}{dx} = -\sin x$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{t} - [\tan x - x] + C = \frac{1}{\cos x} - \tan x + x + C$    

  (d) $\displaystyle \int \frac{1}{\sqrt{1 - x^2 + \sin^{-1}x - x^2\sin^{-1}x}} dx$   (Maple cannot evaluate; Mathematica can.)    
  $\displaystyle \hspace{0.4in} = \int \frac{1}{\sqrt{(1 - x^2)(1 + \sin^{-1}x)}}\...
... = 1 + \sin^{-1}x$,\: $\frac{dt}{dx} = \frac{1}{\sqrt{1 - x^2}}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = 2 t^{\frac{1}{2}} + C = 2 \sqrt{1 + \sin^{-1}x} + C$    

  (e) $\displaystyle \int \frac{1}{\sin x + \cos x} dx = \frac{1}{\sqrt{2}}\int \frac{1}{\cos(x - \frac{\pi}{4})} dx$   (Mathematica cannot evaluate; Maple can.)    
  $\displaystyle \hspace{0.4in} = \frac{1}{\sqrt{2}}\int \frac{\cos(x - \frac{\pi}...
...x - \frac{\pi}{4})$,\: $\frac{dt}{dx} = \cos(x - \frac{\pi}{4})$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{\sqrt{2}} \tanh^{-1} t + C = \frac{1}{\sqrt{2}} \tanh^{-1} \sin\left(x - \frac{\pi}{4}\right) + C$    

(A.2).
$ \displaystyle\int y dx = \int y \frac{dx}{dt} dt$

  (a) $\displaystyle \int x^2\sqrt{x + 4} dx = \int (t^2 - 4)^2 t 2t dt \hspace{0.8in}\fbox{(A.2) $t^2 = x + 4$,\: $\frac{dx}{dt} = 2t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = 2 \int (t^6 - 8t^4 + 16t^2) dt = 2 \left(\frac{1}{7}t^7 - \frac{8}{5}t^5 + \frac{16}{3}t^3\right) + C$    
  $\displaystyle \hspace{0.4in} = \frac{2}{7}(x+4)^\frac{7}{2} - \frac{16}{5}(x+4)^\frac{5}{2} + \frac{32}{3}(x+4)^\frac{3}{2} + C$    

  (b) $\displaystyle \int \sqrt{1 - e^x} dx = \int t \frac{2t}{t^2 - 1} dt \hspace{...
...ox{(A.2) $t^2 = 1 - e^x$,\: $\frac{dx}{dt} = \frac{2t}{t^2 - 1}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = 2\int \left[1 - \frac{1}{1 - t^2}\right] dt = 2...
...t[t - \tanh^{-1} t\right] + C = 2\sqrt{1 - e^x} - 2\tanh^{-1}\sqrt{1 - e^x} + C$    

  (c) $\displaystyle \int \frac{1}{x^4 - 16} dx = \frac{1}{8}\int \frac{1}{t^4 - 1} dt \hspace{0.8in}\fbox{(A.2) $x = 2t$,\: $\frac{dx}{dt} = 2$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{16}\int \left[\frac{1}{t^2 - 1} - \frac{1}{t^2 + 1}\right] dx = \frac{1}{16}\left[-\tanh^{-1}t - \tan^{-1}t\right] + C$    
  $\displaystyle \hspace{0.4in} = -\frac{1}{16}\tanh^{-1}\left(\frac{x}{2}\right) - \frac{1}{16}\tan^{-1}\left(\frac{x}{2}\right) + C$    

  (d) $\displaystyle \int \frac{x^2}{\sqrt{x-1}} dx = 2\int (t^2 + 1)^2 dt \hspace{0.8in}\fbox{(A.2) $t^2 = x - 1$,\: $\frac{dx}{dt} = 2t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = 2\int [t^4 + 2t^2 + 1] dt = \frac{2}{5}t^5 + \f...
...{2}{5}(x-1)^\frac{5}{2} + \frac{4}{3}(x-1)^\frac{3}{2} + 2(x-1)^\frac{1}{2} + C$    

(A.3).
$ \displaystyle\int \frac{1}{y} \frac{dy}{dx} dx = \ln\vert y\vert + C$

  (a) $\displaystyle \int \sec x  dx = \int \sec x  \frac{\sec x + \tan x}{\sec x + ...
...\sec x + \tan x$, \: $\frac{dy}{dx} = \sec x \tan x + \sec^2 x$} \hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \ln \vert y\vert + C = \ln \vert\sec x + \tan x\vert + C$    

(A.4).
Trigonometric substitution.


Substitution   Identities  
       
$ x = a \sin t$, $ \displaystyle t = \sin^{-1} \left(\frac{x}{a}\right)$; $ a^2 - x^2 = a^2 \cos^2 t$, $ dx = a \cos t  dt$.
$ x = a \tan t$ $ \displaystyle t = \tan^{-1} \left(\frac{x}{a}\right)$; $ a^2 + x^2 = a^2 \sec^2 t$, $ dx = a \sec^2 t  dt$.
$ x = a \sec t$ $ \displaystyle t = \cos^{-1} \left(\frac{a}{x}\right)$; $ x^2 - a^2 = a^2 \tan^2 t$, $ dx = a \sec t \tan t  dt$.

  (a) $\displaystyle \displaystyle\int \frac{1}{\sqrt{1 - x^2}} dx = \int dt = t + C = \sin^{-1} x + C \hspace{0.8in}\fbox{(A.4) $x = \sin t$}\hspace{5.0in}$    
  (b) $\displaystyle \displaystyle\int \frac{1}{1 - x^2} dx = \int \sec t dt \hspace{0.8in}\fbox{(A.4) $x = \sin t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \displaystyle\ln \vert\sec t + \tan t\vert + C = \frac{1}{2}\ln\left\vert\frac{1 + x}{1 - x}\right\vert = \tanh^{-1} x + C$    
  (c) $\displaystyle \displaystyle\int \frac{1}{1 + x^2} dx = \int dt = t + C = \tan^{-1} x + C \hspace{0.8in}\fbox{(A.4) $x = \tan t$}\hspace{5.0in}$    
  (d) $\displaystyle \displaystyle\int \frac{1}{\sqrt{1 + x^2}} dx = \int \sec t dt \hspace{0.8in}\fbox{(A.4) $x = \tan t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \displaystyle\ln \vert\sec t + \tan t\vert + C = \ln\left\vert\sqrt{1 + x^2} + x\right\vert = \sinh^{-1} x + C$    
  (e) $\displaystyle \displaystyle\int \frac{1}{\sqrt{x^2 - 1}} dx = \int \sec t dt \hspace{0.8in}\fbox{(A.4) $x = \sec t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \displaystyle\ln \vert\sec t + \tan t\vert + C = \ln\left\vert x + \sqrt{x^2 - 1}\right\vert = \cosh^{-1} x + C$    



Department of Mathematics
Last modified: 2005-10-05