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

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

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

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

2

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= 2 [tan (t)]_{0}^{\frac{π}{4}}

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

f (x) = \frac{3 x}{1 - x}

x \to 0 lim (\frac{7}{x})

- \frac{n ( - 1 ) ^{n}}{n + 1}

\int sin^{2} (x) \cdot cos^{2} (x) d x

f (x) = \frac{1}{x ^{6}}, (2, \frac{1}{64})

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