What is the integral from 0 to pi/2 of 1/(3+cos(2x)) ?

The integral from 0 to pi/2 of 1/(3+cos(2x)) is pi/(4sqrt(2))

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integral from 0 to pi/2 of 1/(3+cos(2x))

\int_{0}^{\frac{π}{2}} \frac{1}{3 + cos ( 2 x )} d x

\frac{π}{4 2}

Decimal

0.55536 \dots

Solution steps

\int_{0}^{\frac{π}{2}} \frac{1}{3 + cos ( 2 x )} d x

Compute the indefinite integral:

\int \frac{1}{3 + cos ( 2 x )} d x = \frac{1}{2 2} arctan (\frac{tan ( x )}{2}) + C

Compute the boundaries:

\int_{0}^{\frac{π}{2}} \frac{1}{3 + cos ( 2 x )} d x = \frac{π}{4 2} - 0

= \frac{π}{4 2} - 0

Simplify

= \frac{π}{4 2}

What is the integral from 0 to pi/2 of 1/(3+cos(2x)) ?
The integral from 0 to pi/2 of 1/(3+cos(2x)) is pi/(4sqrt(2))