What is the integral from 0 to pi/4 of 9sec^2(x) ?

The integral from 0 to pi/4 of 9sec^2(x) is 9

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integral from 0 to pi/4 of 9sec^2(x)

\int_{0}^{\frac{π}{4}} 9 sec^{2} (x) d x

9

Solution steps

\int_{0}^{\frac{π}{4}} 9 sec^{2} (x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{\frac{π}{4}} sec^{2} (x) d x

Use the common integral:

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

= 9 [tan (x)]_{0}^{\frac{π}{4}}

Compute the boundaries:

1

= 9 \cdot 1

Simplify

= 9

x d y + (x - y) d x = 0

\int cos^{8} (x) tan^{3} (x) d x

\int θ d θ

\int_{1}^{4} \frac{x}{2} + \frac{1}{2 x} d x

f (x) = e^{\frac{x}{2}}

What is the integral from 0 to pi/4 of 9sec^2(x) ?
The integral from 0 to pi/4 of 9sec^2(x) is 9