What is the integral from 0 to (7pi)/4 of sec^4(t/7) ?

The integral from 0 to (7pi)/4 of sec^4(t/7) is 28/3

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integral from 0 to (7pi)/4 of sec^4(t/7)

\int_{0}^{\frac{7 π}{4}} sec^{4} (\frac{t}{7}) d t

\frac{28}{3}

Decimal Notation

9.33333 \dots

Solution steps

\int_{0}^{\frac{7 π}{4}} sec^{4} (\frac{t}{7}) d t

Apply u-substitution

: \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 7 d u

= \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 7 d u

Take the constant out:

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

= 7 \cdot \int_{0}^{\frac{π}{4}} sec^{4} (u) d u

Rewrite using trig identities

= 7 \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

= 7 \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

= 7 (\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}

= 7 (1 + \frac{1}{3})

Simplify

7 (1 + \frac{1}{3}) : \frac{28}{3}

= \frac{28}{3}

\int \frac{( 2 x + 5 )}{x ^{2} + 2 x + 5} d x

\int (x^{3} - 3)^{2} d x

\int \frac{cos ( x )}{sin ^{8} ( x )} d x

x^{2} d x + 2 y d y = 0, y (0) = 2

y = \frac{7}{x} (1.7)

What is the integral from 0 to (7pi)/4 of sec^4(t/7) ?
The integral from 0 to (7pi)/4 of sec^4(t/7) is 28/3