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

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

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

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

16

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 4 \cdot 4 [tan (u)]_{0}^{\frac{π}{4}}

Simplify

= 16 [tan (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

1

= 16 \cdot 1

Simplify

= 16

\frac{\partial}{\partial x} (2 x^{3} cos (x y))

x \to 0 lim (\frac{1}{3 ^{x}})

f (x) = x^{3} + 5 x + 7

\frac{4}{s ^{2} + 2 s + 4}

\frac{d}{d x} (\frac{1}{ln ( x + 1 + x ^{2} )})

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