What is the integral from 0 to pi/6 of 17sec^4(x) ?

The integral from 0 to pi/6 of 17sec^4(x) is (170)/(9sqrt(3))

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integral from 0 to pi/6 of 17sec^4(x)

\int_{0}^{\frac{π}{6}} 17 sec^{4} (x) d x

\frac{170}{9 3}

Decimal Notation

10.90550 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 17 sec^{4} (x) d x

Take the constant out:

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

= 17 \cdot \int_{0}^{\frac{π}{6}} sec^{4} (x) d x

Rewrite using trig identities

= 17 \cdot \int_{0}^{\frac{π}{6}} (1 + tan^{2} (x)) sec^{2} (x) d x

Apply u-substitution

: \int_{0}^{\frac{3}{3}} 1 + u^{2} d u

= 17 \cdot \int_{0}^{\frac{3}{3}} 1 + u^{2} 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

= 17 (\int_{0}^{\frac{3}{3}} 1 d u + \int_{0}^{\frac{3}{3}} u^{2} d u)

\int_{0}^{\frac{3}{3}} 1 d u = \frac{1}{3}

\int_{0}^{\frac{3}{3}} u^{2} d u = \frac{1}{9 3}

= 17 (\frac{1}{3} + \frac{1}{9 3})

Simplify

17 (\frac{1}{3} + \frac{1}{9 3}) : \frac{170}{9 3}

= \frac{170}{9 3}

\frac{8 s ^{2} - 4 s + 12}{s ( s ^{2} + 4 )}

f (x) = x, at x = \frac{49}{121}

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

\frac{d ^{4}}{d x ^{4}} (2 e^{x^{2}})

- 3 y

What is the integral from 0 to pi/6 of 17sec^4(x) ?
The integral from 0 to pi/6 of 17sec^4(x) is (170)/(9sqrt(3))