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

The integral from-1 to 1 of 1/(1+sin(x)) is (4tan(1/2))/((tan(1/2)+1)(-tan(1/2)+1))

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integral from-1 to 1 of 1/(1+sin(x))

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

\frac{4 tan ( \frac{1}{2} )}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )}

Decimal

3.11481 \dots

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

Apply the Power Rule

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

Compute the boundaries:

\frac{2}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )} tan (\frac{1}{2})

= 2 \cdot \frac{2}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )} tan (\frac{1}{2})

Simplify

= \frac{4 tan ( \frac{1}{2} )}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )}

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