Question

Evaluate : f (1+tan x tan (x+0)dx​

Answer 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