What is the integral from 0 to pi/2 of 6sec^4(t/2) ?

The integral from 0 to pi/2 of 6sec^4(t/2) is 16

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integral from 0 to pi/2 of 6sec^4(t/2)

\int_{0}^{\frac{π}{2}} 6 sec^{4} (\frac{t}{2}) d t

16

Solution steps

\int_{0}^{\frac{π}{2}} 6 sec^{4} (\frac{t}{2}) d t

Take the constant out:

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

= 6 \cdot \int_{0}^{\frac{π}{2}} sec^{4} (\frac{t}{2}) d t

Apply u-substitution

: \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 2 d u

= 6 \cdot \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 2 d u

Take the constant out:

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

= 6 \cdot 2 \cdot \int_{0}^{\frac{π}{4}} sec^{4} (u) d u

Rewrite using trig identities

= 6 \cdot 2 \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

= 6 \cdot 2 \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

= 6 \cdot 2 (\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}

= 6 \cdot 2 (1 + \frac{1}{3})

Simplify

6 \cdot 2 (1 + \frac{1}{3}) : 16

= 16

\frac{d}{d x} (7 - 3 x^{2})

\frac{\partial}{\partial y} (x^{2} y^{2} z^{2})

\int \frac{ln ( x )}{x ( ln ^{2} ( x ) + 4 )} d x

y^{^{''}} + 36 y = 12 sin (6 t)

\frac{d y}{d x} = \frac{x - 3}{y - 3}

What is the integral from 0 to pi/2 of 6sec^4(t/2) ?
The integral from 0 to pi/2 of 6sec^4(t/2) is 16