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

The integral from 0 to pi/2 of 1/(2cos(x)+3) is (2arctan(1/(sqrt(5))))/(sqrt(5))

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

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

\frac{2 arctan ( \frac{1}{5} )}{5}

Decimal Notation

0.37613 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{2}{u ^{2} + 5} d u

= \int_{0}^{1} \frac{2}{u ^{2} + 5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{1} \frac{1}{u ^{2} + 5} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{\frac{1}{5}} \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{5} \cdot \int_{0}^{\frac{1}{5}} \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{5} [arctan (v)]_{0}^{\frac{1}{5}}

Simplify

2 \cdot \frac{1}{5} [arctan (v)]_{0}^{\frac{1}{5}} : \frac{2}{5} [arctan (v)]_{0}^{\frac{1}{5}}

= \frac{2}{5} [arctan (v)]_{0}^{\frac{1}{5}}

Compute the boundaries:

arctan (\frac{1}{5})

= \frac{2}{5} arctan (\frac{1}{5})

Simplify

= \frac{2 arctan ( \frac{1}{5} )}{5}

\int_{0}^{\frac{1}{2}} \frac{1}{1 - x ^{2}} d x

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\int_{0}^{\frac{1}{2}} x d x

What is the integral from 0 to pi/2 of 1/(2cos(x)+3) ?
The integral from 0 to pi/2 of 1/(2cos(x)+3) is (2arctan(1/(sqrt(5))))/(sqrt(5))