This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
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
 Takeda
 Calculus

Click to edit the document details