What is the integral from 0 to pi/(12) of (sin(6x))/(1+cos^2(6x)) ?

The integral from 0 to pi/(12) of (sin(6x))/(1+cos^2(6x)) is pi/(24)

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integral from 0 to pi/(12) of (sin(6x))/(1+cos^2(6x))

\int_{0}^{\frac{π}{12}} \frac{sin ( 6 x )}{1 + cos ^{2} ( 6 x )} d x

\frac{π}{24}

Decimal Notation

0.13089 \dots

Solution steps

\int_{0}^{\frac{π}{12}} \frac{sin ( 6 x )}{1 + cos ^{2} ( 6 x )} d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} \frac{sin ( u )}{6 ( 1 + cos ^{2} ( u ) )} d u

= \int_{0}^{\frac{π}{2}} \frac{sin ( u )}{6 ( 1 + cos ^{2} ( u ) )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{\frac{π}{2}} \frac{sin ( u )}{1 + cos ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{6} \cdot \int_{1}^{0} - \frac{1}{1 + v ^{2}} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{6} (- \int_{0}^{1} - \frac{1}{1 + v ^{2}} d v)

Take the constant out:

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

= \frac{1}{6} (- (- \int_{0}^{1} \frac{1}{1 + v ^{2}} d v))

Use the common integral:

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

= \frac{1}{6} (- (- [arctan (v)]_{0}^{1}))

Simplify

= \frac{1}{6} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{6} \cdot \frac{π}{4}

Simplify

= \frac{π}{24}

3 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

\frac{d}{d x} (2 sin (x^{2} - 1))

\int \frac{1}{- x ^{2} + 4 x + 12} d x

\frac{x ^{2} + 5 x}{x + 2}

n = 1 \sum \infty \frac{8}{n ^{5}}

What is the integral from 0 to pi/(12) of (sin(6x))/(1+cos^2(6x)) ?
The integral from 0 to pi/(12) of (sin(6x))/(1+cos^2(6x)) is pi/(24)