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

The integral from 0 to pi/4 of 9sec(x)tan(x) is 9(sqrt(2)-1)

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

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

9 (2 - 1)

Decimal Notation

3.72792 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} 1 d u

= 9 \cdot \int_{1}^{2} 1 d u

Integral of a constant:

\int a d x = a x

= 9 [1 \cdot u]_{1}^{2}

Simplify

= 9 [u]_{1}^{2}

Compute the boundaries:

2 - 1

= 9 (2 - 1)

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What is the integral from 0 to pi/4 of 9sec(x)tan(x) ?
The integral from 0 to pi/4 of 9sec(x)tan(x) is 9(sqrt(2)-1)