Math 121 Extra Examples of Trigonometric Substitutions

Extra Examples of Trigonometric Substitutions Math 121 1. Z x2 dx 4 − x2 Recognize difference of squares under the square root Z (2 sin θ)2 p 2 cos θ dθ 4 − 4 sin2 θ Use a sine sub for a2 − x2 . Remember, sub for dx Z 4 sin2 θ p 2 cos θ dθ 4(1 − sin2 θ) Work the algebra to create the identity 1 − sin2 θ = cos2 θ √ = = Z =4 Z =4 Z =4 Z =4 sin2 θ 2 cos θ dθ √ √ 4 cos2 θ The identity creates the perfect square under the root sin2 θ 2 cos θ dθ 2 cos θ Simplify the root of the perfect square, see what cancels sin2 θ dθ Prepare for the EVEN power technique 1 − cos(2θ) dθ 2 Using the half angle identity 1 − cos(2θ) dθ Using a u-sub if needed Z =2   sin(2θ) =2 θ− +C 2   2 sin θ cos θ =2 θ− +C 2 Using the Double Angle Identity = 2 (θ − sin θ cos θ) + C Using the Trig Triangle to unwind the Trig Sub = 2 arcsin x 2 − x √ 2 4 − x2 2 !! + C = 2 arcsin x 2 − ! √ x 4 − x2 +C 2 2 x = 2 sin θ Trig. Substitute dx = 2 cos θdθ x θ √ 4 − x2 2. Try the same problem in a definite integral. You can either change your limits with the trig sub or mark them as x limits. You should choose only one method below. Option 1: Changing Limits √ π 3 Z π 3 (2 sin θ)2 4 sin2 θ p p 2 cos θ dθ 2 cos θ dθ = 2 2 π π 4(1 − sin θ) 1 4 − 4 sin θ 6 6 Z π Z π Z π 2 2 3 3 sin θ 3 sin θ sin2 θ dθ =4 2 cos θ dθ = 4 2 cos θ dθ = 4 √ √ 2 π π π 2 cos θ 4 cos θ 6 6 6   π Z π Z π 3 1 − cos(2θ) 3 sin(2θ) 3 =4 1 − cos(2θ) dθ = 2 θ − dθ = 2 π π 2 2 π 6 6 6       2π  π  √ √ ! sin  π sin 3   π 3 3 π π π 3  = 2  − − = 2 − − + =   3 −  2 6 2 3 4 6 4 3 Z 3 √ x2 dx = 4 − x2 Z Recall: x = 2 sin θ ⇒ θ = arcsin x 2 .   1 π x = 1 ⇒ θ = arcsin = 2 6 Change Limits: x= √ 3 ⇒ θ = arcsin √ ! 3 π = 2 3 Note: If you change your limits, you do not need the Trig Triangle to return to x-variable. Option 2: Mark Limits √ Z 1 3 x2 √ dx = 4 − x2 Z √ x= 3 x=1 (2 sin θ)2 p 2 cos θ dθ = · · · = 2 (θ − sin θ cos θ) 4 − 4 sin2 θ √ x= 3 x=1 using antiderivative from above in 1. ! √ !  √ √ ! √ √    √  3 x 4 − x2 1 1 4−1 3 3 4−3 = 2 arcsin − = 2 arcsin − − 2 arcsin − 2 2 2 2 2 2 1  π  √3  π  √3 π π  π  π 2 − −2 + =2 − =2 = 3 2 6 2 3 6 6 3 x 3. Z 1 5 dx (9 + x2 ) 2 Z 1 2 = 5 3 sec θ dθ √ 9 + 9 tan2 θ 1 2 p 5 3 sec θ dθ Work the algebra to create the identity 1 + tan2 θ = sec2 θ 9(1 + tan2 θ) Z 1 2 √ √ 5 3 sec θ dθ 2 9 sec θ = 3 = 5 3 Z 1 34 Z 1 = 4 3 Z 1 81 Z 1 = 81 Z 1 81 Z = = Use a tangent sub for a2 + x2 . Remember, sub for dx Z = = Recognize sum of squares under the square root The identity creates the perfect square under the root 1 sec2 θ dθ sec5 θ Simplify the root of the perfect square, see what cancels 1 dθ sec3 θ Flip the secant to cosine cos3 θ dθ Prepare for the ODD power technique cos2 θ cos θ dθ Isolate one copy of cosine
1 − sin2 θ cosθ dθ 1 − w2 dw 1 = 81   w3 w− +C 3 1 = 81   sin3 θ sin θ − +C 3 Convert using trig identity Finish with a substitution 1 = 81 √ u = 3 tan θ Trig. Substitute du = 3 sec2 θdθ x 1 √ − 2 3 9+x  x √ 9 + x2 3 ! +C x2 + 9 x θ 3 w = sin θ Standard substitution dw = cos θ dθ Z 4. x2 1 x arcsin x dx = arcsin x − 2 2 Z √ x2 dx 1 − x2 sin2 θ x2 1 = arcsin x − 2 2 Z x2 1 p · cos θ dθ = arcsin x − 2 2 2 1 − sin θ x2 1 = arcsin x − 2 2 Z sin2 θ x2 1 · cos θ dθ = arcsin x − cos θ 2 2 x2 1 = arcsin x − 2 2 Z 1 − cos(2θ) x2 1 dθ = arcsin x − 2 2 4 Z Z sin2 θ √ · cos θ dθ cos2 θ sin2 θdθ Z 1 − cos(2θ)dθ =   x2 1 1 x2 1 1 arcsin x − θ − sin(2θ) + C = arcsin x − θ + sin(2θ) + C 2 4 2 2 4 8 = x2 1 1 x2 1 1 p arcsin x − θ + (2 sin θ cos θ) + C = arcsin x − arcsin x + x 1 − x2 + C 2 4 8 2 4 4 u = arcsin x IBP to start: du = √ dv = xdx 1 x2 dx v = 2 1 − x2 1 x = sin θ Trig. Substitute dx = cos θdθ x θ √ 1 − x2