What is the integral from 0 to pi/4 of 2/(1+sin(2x)) ?

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

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integral from 0 to pi/4 of 2/(1+sin(2x))

\int_{0}^{\frac{π}{4}} \frac{2}{1 + sin ( 2 x )} d x

1

Solution steps

\int_{0}^{\frac{π}{4}} \frac{2}{1 + sin ( 2 x )} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{\frac{π}{4}} \frac{1}{1 + sin ( 2 x )} d x

Apply u-substitution

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

= 2 \cdot \int_{0}^{\frac{π}{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

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

Apply u-substitution

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{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

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

Complete the square

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

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

Apply u-substitution

: \int_{1}^{2} \frac{1}{w ^{2}} d w

= 2 \cdot \frac{1}{2} \cdot 2 \cdot \int_{1}^{2} \frac{1}{w ^{2}} d w

Apply the Power Rule

: [- \frac{1}{w}]_{1}^{2}

= 2 \cdot \frac{1}{2} \cdot 2 [- \frac{1}{w}]_{1}^{2}

Simplify

2 \cdot \frac{1}{2} \cdot 2 [- \frac{1}{w}]_{1}^{2} : 2 [- \frac{1}{w}]_{1}^{2}

= 2 [- \frac{1}{w}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Simplify

= 1

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