What is the integral from 0 to (5pi)/4 of sec^4(t/5) ?

The integral from 0 to (5pi)/4 of sec^4(t/5) is 20/3

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integral from 0 to (5pi)/4 of sec^4(t/5)

\int_{0}^{\frac{5 π}{4}} sec^{4} (\frac{t}{5}) d t

\frac{20}{3}

Decimal Notation

6.66666 \dots

Solution steps

\int_{0}^{\frac{5 π}{4}} sec^{4} (\frac{t}{5}) d t

Apply u-substitution

: \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 5 d u

= \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 5 d u

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{π}{4}} sec^{4} (u) d u

Rewrite using trig identities

= 5 \cdot \int_{0}^{\frac{π}{4}} (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

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

= 5 \cdot \int_{0}^{1} 1 + v^{2} d v

Apply the Sum Rule:

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

= 5 (\int_{0}^{1} 1 d v + \int_{0}^{1} v^{2} d v)

\int_{0}^{1} 1 d v = 1

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

= 5 (1 + \frac{1}{3})

Simplify

5 (1 + \frac{1}{3}) : \frac{20}{3}

= \frac{20}{3}

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What is the integral from 0 to (5pi)/4 of sec^4(t/5) ?
The integral from 0 to (5pi)/4 of sec^4(t/5) is 20/3