By Paul Dawkins
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Additional resources for Calculus II
So, let’s work a couple examples using substitutions and tables. Example 1 Evaluate the following integral. ó 7 + 9x 2 dx ô x2 õ Solution So, the first thing we should do is go to the tables and see if there is anything in the tables that is close to this. In the tables in Stewart we find the following integral, ( ) a2 + u 2 ó a2 + u2 du = + ln u + a 2 + u 2 + c ô 2 u u õ This is nearly what we’ve got in our integral. aspx Calculus II coefficient in front of the x2 and the formula doesn’t. This is easily enough dealt with.
In the tables in Stewart we find the following integral, ( ) a2 + u 2 ó a2 + u2 du = + ln u + a 2 + u 2 + c ô 2 u u õ This is nearly what we’ve got in our integral. aspx Calculus II coefficient in front of the x2 and the formula doesn’t. This is easily enough dealt with. All we need to do is the following manipulation on the integrand. 2 2 ó 9 ( 97 + x ) ó 3 79 + x ó ó 7 + 9x2 dx = ô dx = ô dx = 3ô ô 2 2 2 x x x õ õ õ õ So, we can now use the formula with a = 7 9 + x2 x2 dx 7 . 3 æ ó 7 + 9x2 çdx = 3 ô ç x2 õ è 7 9 + x2 x æ öö 7 + ln çç x + + x 2 ÷÷ ÷ + c 9 è ø ÷ø Example 2 Evaluate the following integral.
From our original substitution we have, sec q = 5 x hypotenuse = 2 adjacent This gives the following right triangle. From this we can see that, tan q = 25 x 2 - 4 2 We can deal with the q in one of any variety of ways. aspx Calculus II æ 5x ö q = sec -1 ç ÷ è 2 ø While this is a perfectly acceptable method of dealing with the q we can use any of the possible six inverse trig functions and since sine and cosine are the two trig functions most people are familiar with we will usually use the inverse sine or inverse cosine.