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Unformatted text preview: (π 2 , π 2 ) and the range to (∞ , ∞ ). Similarly, the domain D 2 of sin1 x is [1 , 1]. The range R 2 is [π 2 , π 2 ]. The function sin1 x is onetoone in the domain D 2 and the range R 2 , and so the inverse exists. Step 3. The range R 1 of tan x must agree with the domain D 2 of sin1 x , i.e., R 1 = D 2 = [1 , 1]. To achieve this we will restrict the domain D 1 of tan x to [π 4 , π 4 ]. Thus, the domain of f ( x ) is [π 4 , π 4 ] and the range is [π 2 , π 2 ]. b. Calculate the inverse function f1 ( x ) and specify its domain. The inverse function f1 ( x ) = tan1 (sin x ). Pulling directly from the range above, the domain is [π 2 , π 2 ]. 4. Evaluate R dx 5 x √ ln(3 x ) . 5. Evaluate ln x x (1+ln x ) dx . 6. Evaluate R √ ln π 2 xe x 2 cos( e x 2 ) dx . 1...
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This note was uploaded on 06/20/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 BERMAN
 Math, Calculus

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