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Question : 75 of 160
Marks:
+1,
-0
Solution:
Given,
f(t)=tan(2n−1)x‌dx,n∈So,
f(t+π)=tan(2n−1)x‌dx=tan(2n−1)x‌dx+tan(2n−1)x‌dxLet
u=x−π⇒du=dxLimit, when
x=t,u=t−π and
x=t+Ï€,u=tSo,
f(t+π)=f(t)+tan(2n−1)(u+π)duSince,
tan(u+π)=tan‌u‌=f(t)+tan(2n−1(u)du‌=f(t)+tan(2n−1)(u)du+tan(2n−1)(u)duLet
v=−u, the
dv=−duWhen
u=t−π,v=π−t, when
u=0,v=0, so we get
‌=f(t)−tan(2n−1)(−v)dv+tan(2n−1)(u)du‌‌‌=f(t)+tan(2n−1)(v)dv+tan(2n−1)(u)du‌=f(t)+tan(2n−1)(v)dv+f(t)‌=2f(t)−tan(2n−1)(v)dv‌=2f(t)−tan(2n−1)(v)dv−tan(2n−1)(v)dvLet
w=v−π, then
dw=dvWhen
v=π−t,w=−t, when
v=πw=0‌=2f(t)−tan(2n−1)(v)dv−tan(2n−1)(w+π)dw‌2f(t)−tan(2n−1)(v)dv−tan(2n−1)(w)dwLet
z=−w, then
dz=−dw when
w=−t,z=t and
w=0,z=0=‌2f(t)−tan(2n−1)(v)dv+tan(2n−1)(−z)dz=‌2f(t)−tan(2n−1)(v)dv−tan(2n−1)(z)dz=‌2f(t)−tan(2n−1)(v)dv−f(t)=‌f(t)−tan(2n−1)(v)dv=‌f(t)+0[∵tan(2n−1)(v)dv=0]‌=f(t)+f(π)‌‌[∵f(π)=0]∴‌f(t+π)=f(t)+f(π)
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