ma1201_assignment2_sol - MA 1201 Semester B 2018\/19 Assignment 2 \u2014 Due at 5 pm(Thursday in tutorial classes Instructions \u2022 Please show your work

# ma1201_assignment2_sol - MA 1201 Semester B 2018/19...

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MA 1201 Semester B 2018/19 Assignment 2 — Due at 5 pm, 21/3/2019 (Thursday) in tutorial classes Instructions: Please show your work. Unsupported answers will receive NO credits. Make sure you write down the correct lecture session (A/B/C/D/E/F/G/H) you have registered for, together with your full name and student ID on the front page of your answer script. Hand your solution to your TA in tutorial class before the deadline. In case your tutorial class is scheduled on days other than Thursday, you may choose to hand your solution to your TA in one of the tutorial classes scheduled on Thursday. NO late homework will be accepted. Homework submitted to wrong tutorial sessions will NOT be graded and will receive 0 POINTS . 1. (50 points) Evaluate the following integrals. (a) (10 points) Z ( sin - 1 x ) 2 dx . Solution . In view of the presence of the inverse trig function, the idea is to apply integration by parts. Let (see page 81 of the supplementary notes for the ILATE rule for choosing u ) u = ( sin - 1 x ) 2 , dv = dx . Direct integration yields v = Z dv = Z dx = x , so Z ( sin - 1 x ) 2 dx = Z ( sin - 1 x ) 2 d ( x ) = ( sin - 1 x ) 2 · x - Z xd ( sin - 1 x ) 2 = x ( sin - 1 x ) 2 - Z x · 2sin - 1 x · 1 1 - x 2 dx = x ( sin - 1 x ) 2 - 2 Z x 1 - x 2 sin - 1 xdx . For the integral on the right side, apply integration by parts one more time with u = sin - 1 x , dv = x 1 - x 2 dx , v = Z dv = Z x 1 - x 2 dx = Z 1 1 - x 2 · xdx w = 1 - x 2 = dw = - 2 xdx Z w - 1 / 2 - 1 2 dw = - 1 2 · 2 w 1 / 2 = - p 1 - x 2 , to deduce Z x 1 - x 2 sin - 1 xdx = Z sin - 1 x · x 1 - x 2 dx = Z sin - 1 xd ( - p 1 - x 2 ) = sin - 1 x · ( - p 1 - x 2 ) - Z ( - p 1 - x 2 ) d ( sin - 1 x ) = - p 1 - x 2 sin - 1 x + Z p 1 - x 2 · 1 1 - x 2 dx = - p 1 - x 2 sin - 1 x + Z 1 dx = - p 1 - x 2 sin - 1 x + x + C . 1
Thus Z ( sin - 1 x ) 2 dx = x ( sin - 1 x ) 2 - 2 ( - p 1 - x 2 sin - 1 x + x ) + C = x ( sin - 1 x ) 2 + 2 p 1 - x 2 sin - 1 x - 2 x + C . Remark . An alternative way to solve the problem is to introduce the (trig) substitution x = sin θ , - π 2 θ π 2 , so that sin - 1 x = sin - 1 ( sin θ ) = θ , dx = ( sin θ ) 0 d θ = cos θ d θ . This leads to a simpler integral Z ( sin - 1 x ) 2 dx = Z θ 2 · ( cos θ d θ ) , to which the (repeated) integration by parts is applied more easily: Z θ 2 cos θ d θ = Z θ 2 d ( sin θ ) = θ 2 · sin θ - Z sin θ d ( θ 2 ) = θ 2 sin θ - 2 Z θ sin θ d θ = θ 2 sin θ + 2 Z θ d ( cos θ ) = θ 2 sin θ + 2 θ · cos θ - Z cos θ d θ = θ 2 sin θ + 2 θ cos θ - 2sin θ + C . (b) (10 points) Z sin3 x cos 2 2 xdx . Motivation . The even power cos 2 2 x needs to be simplified using the half-angle formula: cos 2 2 x = 1 2 1 + cos2 ( 2 x ) = 1 2 ( 1 + cos4 x ) , and the resulting product with sin3 x needs to be simplified using the product-to-sum formula: sin3 x cos4 x = 1 2 sin ( 3 x + 4 x )+ sin ( 3 x - 4 x ) = 1 2 sin ( 7 x )+ sin ( - x ) = 1 2 ( sin7 x - sin x ) . Solution . Z sin3 x cos 2 2 xdx = Z sin3 x · 1 2 ( 1 + cos4 x ) dx = 1 2 Z ( sin3 x + sin3 x cos4 x ) dx = 1 2 Z sin3 xdx + 1 2 Z sin3 x cos4 xdx = 1 2 Z sin3 xdx + 1 2 Z 1 2 ( sin7 x - sin x ) dx = 1 2 · - 1 3 cos3 x + 1 4 · - 1 7 cos7 x + cos x + C = - 1 6 cos3 x - 1 28 cos7 x + 1 4 cos x + C .
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