What is the value of 2(1)-2arctan(1)-ln(1^2+1) ?

The value of 2(1)-2arctan(1)-ln(1^2+1) is 2-pi/2-ln(2)

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2(1)-2arctan(1)-ln(1^2+1)

2 (1) - 2 arctan (1) - ln (1^{2} + 1)

2 - \frac{π}{2} - ln (2)

Decimal Notation

- 0.26394 \dots

Solution steps

2 (1) - 2 arctan (1) - ln (1^{2} + 1)

= 2 \cdot 1 - 2 arctan (1) - ln (1^{2} + 1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= 2 \cdot 1 - 2 \cdot \frac{π}{4} - ln (1^{2} + 1)

Simplify

2 \cdot 1 - 2 \cdot \frac{π}{4} - ln (1^{2} + 1) : 2 - \frac{π}{2} - ln (2)

2 \cdot 1 - 2 \cdot \frac{π}{4} - ln (1^{2} + 1)

Apply rule

1^{a} = 1

1^{2} = 1

= 2 \cdot 1 - 2 \cdot \frac{π}{4} - ln (1 + 1)

2 \cdot 1 = 2

2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{4}

Cancel the common factor:

2

= \frac{π}{2}

ln (1 + 1) = ln (2)

ln (1 + 1)

Add the numbers:

1 + 1 = 2

= ln (2)

= 2 - \frac{π}{2} - ln (2)

= 2 - \frac{π}{2} - ln (2)

arctan (\frac{6 3}{6})

arcsin (\frac{1}{2.2})

2.5 \cdot cos (3 0^{\circ})

sin (9 0^{\circ}) - sin (53.1 3^{\circ})

cos^{2} (\frac{1}{2} \cdot \frac{2 π}{3})

What is the value of 2(1)-2arctan(1)-ln(1^2+1) ?
The value of 2(1)-2arctan(1)-ln(1^2+1) is 2-pi/2-ln(2)