Evaluate : f (1+tan x tan (x+0)dx
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1
Answer:
Given
f
′
(x)=tan
−1
(secx+tanx)
=tan
−1
(
cosx
1+sinx
)
=tan
−1
⎝
⎛
sin(
2
π
+x)
1−cos(
2
π
+x)
⎠
⎞
=tan
−1
⎝
⎛
2sin(
4
π
+
2
x
)cos(
4
π
+
2
x
)
2sin
2
(
4
π
+
2
x
)
⎠
⎞
=tan
−1
(tan(
4
π
+
2
x
))
=
4
π
+
2
x
∫(f
′
(x))dx=∫(
4
π
+
2
x
)dx
f(x)=
4
π
x+
4
x
2
+c
f(0)=c=0⇒f(x)=
4
π
x+
4
x
2
So, f(1)=
4
π+1
∴
f(1)=
4
π+1
....Answer
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