MAT3CZ - Assignment 4 Solutions: Integral Calculus, Semester 1

This document presents solutions to several integral problems, including contour plots and the application of the tangent half-angle substitution method. It also addresses integrals with singularities and calculates Cauchy's principal value. Furthermore, it explores Fourier transforms of given functions, providing step-by-step evaluations and MATLAB code for generating contour plots. The assignment covers topics from complex analysis to real integrals.

Running head: INTEGRAL
INTEGRAL
Name of the Student
Name of the University
Author Note

1INTEGRAL
4.
The given integral is
I = ∫
0
2 π
1
cos ( θ ) +2 sin ( θ ) +3 dθ
The contour of the function is a polar plot as θ is real and in radians. There are no singular
points of the function in the range 0 to 2 π or the function is not zero anywhere within
specified range.
MATLAB code for contour plot:
theta = linspace(0,360,50);
func = 1./(cos(theta) + 2.*sin(theta) + 3);
theta_radians = deg2rad(theta);
polarplot(theta_radians,func)
grid on
Contour Plot:

2INTEGRAL
0
30
60
90
120
150
180
210
240
270
300
330
0
0.5
1
I = ∫
0
2 π
1
cos ( θ ) +2 sin ( θ ) +3 dθ
By, tangent half-angle substitution method
I = ∫ 1
¿ ¿ ¿ dθ
Substituting u = tan( θ/2)
 dt= 2
sec2
( θ
2 ) du= 2
u2+1 du
Hence, I = ∫ 1
u2 +2 u+2 du = ∫ 1
( u+1 ) 2 +1 du = ∫ 1
v2 +1 dv = atan(u+1)
= [atan(tan(θ /2)+1)]0
2 π
= π.

3INTEGRAL
5.
The given integral is
I =∫
−∞
∞
1
x3−x2 + x dx
Now, the integral has singularities at points x1 = 0, x2 = 0.5 + 0.866i, x3 = 0.5 – 0.866i.
As, only one real singularity point is at x = 0.
Hence, the Cauchy principle value of the integral I will be
I = lim ¿e →0 ¿ ¿
Evaluating these integrals gives,
I = -(ln(e^2 + e + 1)/2) – atan
( 2 √ 3 e+ √ 3 )
3
√ 3 + π
2 √ 3 +1 + (ln(e^2 + e + 1)/2) - atan
( 2 √ 3 e+ √ 3 )
3
√ 3
+
π
2 √ 3
= π
√ 3
MATLAB code for contour plot:
x = -10000:1:10000;
func = 1./(x.^3 - x.^2 + x);
polarplot(x,func)
Plot:

4INTEGRAL
0
30
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90
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270
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330
0
0.2
0.4
0.6
0.8
1
With the expansion of the contour the integral of the contour approaches to zero as the only
discontinuity is at x= 0 and the value of the closed path evaluates to zero.
6.
a) f ( t )= 1
t2+2+2
Fourier transform of the function is given by,
F ( t )=∫
−∞
∞
1
t2 +2+2 exp (−iwt ) dt
Evaluating the integral we get F(t) in terms of w
F(t) = √π /6 exp ( w− √3 w ) (when w>0).
b) ^f ( w ) = 1
( w +i ) 2 ( w−i )

5INTEGRAL
The inverse Fourier transform is given by,
f(t) = F−1 ( ^f ( w ) )
= 1
2 π ∫
−∞
∞
1
( w+i )2 ( w−i ) exp ( iwt ) dw
Evaluating the integral and assuming t>0 we get,
f(t) = ( −1
2 ) i √ π
2 exp ( −t ) ( 2 t+1 )