What is the integral from pi/6 to pi/3 of 7csc^3(x) ?

The integral from pi/6 to pi/3 of 7csc^3(x) is (7(-3ln(3(7-4sqrt(3)))+12sqrt(3)-4))/(12)

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integral from pi/6 to pi/3 of 7csc^3(x)

\int_{\frac{π}{6}}^{\frac{π}{3}} 7 csc^{3} (x) d x

\frac{7 ( - 3 ln ( 3 ( 7 - 4 3 ) ) + 12 3 - 4 )}{12}

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{3}} 7 csc^{3} (x) d x

Take the constant out:

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

= 7 \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} csc^{3} (x) d x

Apply u-substitution

: \int_{2 - 3}^{\frac{3}{3}} \frac{( 1 + u ^{2} ) ^{2}}{4 u ^{3}} d u

= 7 \cdot \int_{2 - 3}^{\frac{3}{3}} \frac{( 1 + u ^{2} ) ^{2}}{4 u ^{3}} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int_{2 - 3}^{\frac{3}{3}} \frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} d u

Expand

\frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} : \frac{1}{u ^{3}} + \frac{2}{u} + u

= 7 \cdot \frac{1}{4} \cdot \int_{2 - 3}^{\frac{3}{3}} \frac{1}{u ^{3}} + \frac{2}{u} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 \cdot \frac{1}{4} (\int_{2 - 3}^{\frac{3}{3}} \frac{1}{u ^{3}} d u + \int_{2 - 3}^{\frac{3}{3}} \frac{2}{u} d u + \int_{2 - 3}^{\frac{3}{3}} u d u)

\int_{2 - 3}^{\frac{3}{3}} \frac{1}{u ^{3}} d u = 2 (1 + 3)

\int_{2 - 3}^{\frac{3}{3}} \frac{2}{u} d u = - ln (3 (7 - 43))

\int_{2 - 3}^{\frac{3}{3}} u d u = \frac{1}{6} - \frac{7 - 4 3}{2}

= 7 \cdot \frac{1}{4} (2 (1 + 3) - ln (3 (7 - 43)) + \frac{1}{6} - \frac{7 - 4 3}{2})

Simplify

7 \cdot \frac{1}{4} (2 (1 + 3) - ln (3 (7 - 43)) + \frac{1}{6} - \frac{7 - 4 3}{2}) : \frac{7 ( - 3 ln ( 3 ( 7 - 4 3 ) ) + 12 3 - 4 )}{12}

= \frac{7 ( - 3 ln ( 3 ( 7 - 4 3 ) ) + 12 3 - 4 )}{12}

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What is the integral from pi/6 to pi/3 of 7csc^3(x) ?
The integral from pi/6 to pi/3 of 7csc^3(x) is (7(-3ln(3(7-4sqrt(3)))+12sqrt(3)-4))/(12)