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Unformatted text preview: 5.3. TRIGONOMETRIC SUBSTITUTIONS 47 5.3 Trigonometric Substitutions Return more midterms? Rough meaning of grades: 29–34 is A 23–28 is B 17–22 is C 11–16 is D Regarding the quiz—if you do every homework problem that was assigned, you’ll have a severe case of deja vu on the quiz! On the exam, we do not restrict ourselves like this, but you get to have a sheet of paper. The first homework problem is to compute Z 2 √ 2 1 x 3 √ x 2- 1 dx. (5.3.1) Your first idea might be to do some sort of substitution, e.g., u = x 2- 1, but du = 2 xdx is nowhere to be seen and this simply doesn’t work. Likewise, integration by parts gets us nowhere. However, a technique called “inverse trig substitutions” and a trig identity easily dispenses with the above integral and several similar ones! Here’s the crucial table: Expression Inverse Substitution Relevant Trig Identity √ a 2- x 2 x = a sin( θ ) ,- π 2 ≤ θ ≤ π 2 1- sin 2 ( θ ) = cos 2 ( θ ) √ a 2 + x 2 x = a tan( θ ) ,- π 2 < θ < π 2 1 + tan 2 ( θ ) = sec 2 ( θ ) √ x 2- a 2 x = a sec( θ ) , ≤ θ < π 2 or π ≤ θ < 3 π 2 sec 2 ( θ )- 1 = tan 2 ( θ ) Inverse substitution works as follows. If we write x = g ( t ), then Z f ( x ) dx = Z f ( g ( t )) g ( t ) dt....
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This note was uploaded on 12/26/2009 for the course MATH MATH 20B taught by Professor Takeda during the Winter '07 term at UCSD.
- Winter '07