Trigonometric Integrals (7.2)

# Trigonometric Integrals (7.2) - 2-4 The square is complete...

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Trigonometric integrals (7.2) Integral u = du = Identity to use Works when R sin m x cos n xdx sin x cos xdx sin 2 x + cos 2 x = 1 n is odd cos x - sin xdx m is odd R tan m x sec n xdx tan x sec 2 xdx tan 2 x + 1 = sec 2 x n is even sec x sec x tan xdx m is odd If both sin x and cos x appear with even exponents, use sin 2 x = 1 - cos2 x 2 cos 2 x = 1 + cos2 x 2 sin x cos x = sin2 x 2 Trigonometric substitutions (7.3) Integral has x = ... = dx = a 2 - x 2 a sin θ a cos θ a cos θ dθ a 2 + x 2 a tan θ a sec θ a sec 2 θ dθ x 2 - a 2 a sec θ a tan θ a sec θ tan θ dθ If ··· is multiplied by x to an odd power, use u = ··· instead of trig sub. How to complete the square Given: x 2 + 6 x + 5 Let b be the coeﬃcient of x divided by 2. In this example b = 6 / 2 = 3. Replace the two terms with x by ( x + b ) 2 - b 2 . Here I replace x 2 + 6 x with ( x + 3) 2 - 9. The result: ( x + 3) 2 - 9 + 5, which simpliﬁes to ( x + 3)
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Unformatted text preview: 2-4. The square is complete! • (if given 3 x 2 + 18 x + 15 , factor out 3 to get 3( x 2 + 6 x + 5) , then proceed as above) If this was a trig substitution problem, you would sub x + 3 = 2sec θ . In a partial fraction problem you want to know if the quadratic is reducible or not. If after completing the square you have a sum, then the quadratic is irreducible. If you have a diﬀerence, like ( x + 3) 2-4 in my example, then it’s reducible. Use the formula a 2-b 2 = ( a-b )( a + b ) to reduce it: ( x + 3) 2-4 = ( x + 3-2)( x + 3 + 2) = ( x + 1)( x + 5)....
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