What is the integral from sqrt(pi/(12)) to sqrt(pi/9 of)5xtan^4(3x^2)sec^2(3x^2) ?

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integral from sqrt(pi/(12)) to sqrt(pi/9 of)5xtan^4(3x^2)sec^2(3x^2)

\int_{\frac{π}{12}}^{\frac{π}{9}} 5 x tan^{4} (3 x^{2}) sec^{2} (3 x^{2}) d x

\frac{9 3 - 1}{6}

Decimal

2.43140 \dots

Solution steps

\int_{\frac{π}{12}}^{\frac{π}{9}} 5 x tan^{4} (3 x^{2}) sec^{2} (3 x^{2}) d x

Take the constant out:

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

= 5 \cdot \int_{\frac{π}{12}}^{\frac{π}{9}} x tan^{4} (3 x^{2}) sec^{2} (3 x^{2}) d x

Apply u-substitution

: \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{tan ^{4} ( u ) sec ^{2} ( u )}{6} d u

= 5 \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{tan ^{4} ( u ) sec ^{2} ( u )}{6} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} tan^{4} (u) sec^{2} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{6} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} (1 + tan^{2} (u)) tan^{4} (u) d u

Apply u-substitution

: \int_{1}^{3} v^{4} d v

= 5 \cdot \frac{1}{6} \cdot \int_{1}^{3} v^{4} d v

Apply the Power Rule

: [\frac{v ^{5}}{5}]_{1}^{3}

= 5 \cdot \frac{1}{6} [\frac{v ^{5}}{5}]_{1}^{3}

Simplify

5 \cdot \frac{1}{6} [\frac{v ^{5}}{5}]_{1}^{3} : \frac{5}{6} [\frac{v ^{5}}{5}]_{1}^{3}

= \frac{5}{6} [\frac{v ^{5}}{5}]_{1}^{3}

Compute the boundaries:

\frac{9 3}{5} - \frac{1}{5}

= \frac{5}{6} (\frac{9 3}{5} - \frac{1}{5})

Simplify

= \frac{9 3 - 1}{6}

\int_{0}^{1} x^{\frac{5}{6}} d x

\int_{0}^{1} (x + 2 1 - x^{2}) d x

\int_{0}^{1} \frac{1}{4 + 5 x - 1} d x

\int_{0}^{5} \frac{x}{x ^{2} + 1} d x

\int_{4}^{5} 1

What is the integral from sqrt(pi/(12)) to sqrt(pi/9 of)5xtan^4(3x^2)sec^2(3x^2) ?
The integral from sqrt(pi/(12)) to sqrt(pi/9 of)5xtan^4(3x^2)sec^2(3x^2) is (9sqrt(3)-1)/6