What is the integral from 0 to 1.5 of 1/(1+sin(x)) ?

The integral from 0 to 1.5 of 1/(1+sin(x)) is 0.96458…

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integral from 0 to 1.5 of 1/(1+sin(x))

\int_{0}^{1.5} \frac{1}{1 + sin ( x )} d x

0.96458 \dots

Solution steps

\int_{0}^{1.5} \frac{1}{1 + sin ( x )} d x

Apply u-substitution

: \int_{0}^{0.93159 \dots} \frac{2}{1 + u ^{2} + 2 u} d u

= \int_{0}^{0.93159 \dots} \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_{0}^{0.93159 \dots} \frac{1}{1 + u ^{2} + 2 u} d u

Complete the square

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

= 2 \cdot \int_{0}^{0.93159 \dots} \frac{1}{( u + 1 ) ^{2}} d u

Apply u-substitution

: \int_{1}^{1.93159 \dots} \frac{1}{v ^{2}} d v

= 2 \cdot \int_{1}^{1.93159 \dots} \frac{1}{v ^{2}} d v

Apply the Power Rule

: [- \frac{1}{v}]_{1}^{1.93159 \dots}

= 2 [- \frac{1}{v}]_{1}^{1.93159 \dots}

Compute the boundaries:

0.48229 \dots

= 2 \cdot 0.48229 \dots

Simplify

= 0.96458 \dots

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What is the integral from 0 to 1.5 of 1/(1+sin(x)) ?
The integral from 0 to 1.5 of 1/(1+sin(x)) is 0.96458…