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. ϕ ( π ) = 1, ϕ ( x)<br />
<br />
y<br />
∫ [sin x −ϕ ( x)]<br />
dx + ϕ ( x)<br />
dy .<br />
x<br />
L<br />
P<br />
Q<br />
∂P<br />
∂Q<br />
=<br />
∂y<br />
∂x<br />
1<br />
[sin x − ϕ ( x)]<br />
= ϕ′<br />
( x)<br />
x<br />
<br />
1 sinx<br />
ϕ′ ( x)<br />
+ ϕ(<br />
x)<br />
= , ϕ(<br />
π ) =<br />
x x<br />
ϕ ( x)<br />
=<br />
1<br />
?