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Mathematics
Mathematics
Problem
(a) Show
∫
1
1
+
cos
θ
d
θ
=
k
tan
θ
2
+
C
\int \frac{1}{1+\cos\theta} d\theta = k \tan \frac{\theta}{2} + C
∫
1
+
c
o
s
θ
1
d
θ
=
k
tan
2
θ
+
C
, where
k
k
k
is a constant to be determined. (b) Hence find the general solution of the differential equation
(
1
+
cos
θ
)
d
x
d
θ
+
x
2
=
1.
(1 + \cos \theta) \frac{dx}{d\theta} + x^2 = 1.
(
1
+
cos
θ
)
d
θ
d
x
+
x
2
=
1.
[There is no need to simplify your general solution.]
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