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

The integral from 0 to 1 of 1/(1+sin(2x)) is -1/(tan(1)+1)+1

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integral from 0 to 1 of 1/(1+sin(2x))

\int_{0}^{1} \frac{1}{1 + sin ( 2 x )} d x

- \frac{1}{tan ( 1 ) + 1} + 1

Decimal

0.60897 \dots

Solution steps

\int_{0}^{1} \frac{1}{1 + sin ( 2 x )} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{2 ( 1 + sin ( u ) )} d u

= \int_{0}^{2} \frac{1}{2 ( 1 + sin ( u ) )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2} \frac{1}{1 + sin ( u )} d u

Apply u-substitution

: \int_{0}^{t a n (1)} \frac{2}{1 + v ^{2} + 2 v} d v

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

Take the constant out:

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

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

Complete the square

1 + v^{2} + 2 v : (v + 1)^{2}

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

Apply u-substitution

: \int_{1}^{t a n (1) + 1} \frac{1}{w ^{2}} d w

= \frac{1}{2} \cdot 2 \cdot \int_{1}^{t a n (1) + 1} \frac{1}{w ^{2}} d w

Apply the Power Rule

: [- \frac{1}{w}]_{1}^{t a n (1) + 1}

= \frac{1}{2} \cdot 2 [- \frac{1}{w}]_{1}^{t a n (1) + 1}

Simplify

\frac{1}{2} \cdot 2 [- \frac{1}{w}]_{1}^{t a n (1) + 1} : [- \frac{1}{w}]_{1}^{t a n (1) + 1}

= [- \frac{1}{w}]_{1}^{t a n (1) + 1}

Compute the boundaries:

- \frac{1}{tan ( 1 ) + 1} + 1

= - \frac{1}{tan ( 1 ) + 1} + 1

\int \frac{( x + 6 )}{( x + 7 )} d x

\frac{d}{d y} (2 x - y)

\frac{d}{d x} (\frac{ln ( x ^{2} )}{x ^{2}})

\int \frac{1}{9 - 2 x} d x

x \to \infty lim (- \frac{2}{x})

What is the integral from 0 to 1 of 1/(1+sin(2x)) ?
The integral from 0 to 1 of 1/(1+sin(2x)) is -1/(tan(1)+1)+1