Question

integrations of 1/1+cosx from pi/4to 3Pi/4 equal to 2

Answer 1

$\int\frac{dx}{1+cosx} \\ \\ = \int \frac{dx}{2 {cos}^{2} \frac{x}{2} } \\ \\ = \int \frac{1}{2} {sec}^{2} \frac{x}{2} dx \\ \\ = \frac{1}{2} \int {sec}^{2} \frac{x}{2} dx$
so, I = 1/2 [ tanx/2]/(1/2) = [ tanx/2]
now put limit
I = [tan3π/8 - tanπ/8]

$\text{we know}$
$\boxed{\boxed{\bold{tanx=\sqrt{\frac{1-cos2x}{1+cos2x}}}}}$


tanπ/8 = tan22.5° = √{(1 - cos45°)/(1+sin45°)}
= √{ (1 - 1/√2)/(1 + 1/√2)}
= √{ (√2 - 1)/(√2 + 1)}
= √2 - 1
so, tan(3π/8) = tan{π/2 -π/8}=cotπ/8 = 1/(√2-1)
= √2 + 1

so,
I = [ tan3π/8 - tanπ/8]
= [ √2 + 1 - √2 + 1 ] = 2

$\text{hence, answer is }\boxed{\boxed{\bold{2}}}$