What is the integral from 0 to pi/4 of 3sec(x)tan(x) ?

The integral from 0 to pi/4 of 3sec(x)tan(x) is 3(sqrt(2)-1)

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integral from 0 to pi/4 of 3sec(x)tan(x)

\int_{0}^{\frac{π}{4}} 3 sec (x) tan (x) d x

3 (2 - 1)

Solution steps

\int_{0}^{\frac{π}{4}} 3 sec (x) tan (x) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{\frac{π}{4}} sec (x) tan (x) d x

Apply u-substitution

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

= 3 \cdot \int_{1}^{2} 1 d u

Integral of a constant:

\int a d x = a x

= 3 [1 \cdot u]_{1}^{2}

Simplify

= 3 [u]_{1}^{2}

Compute the boundaries:

2 - 1

= 3 (2 - 1)

\int_{2}^{4} \frac{1}{( x - 2 ) ^{2}} d x

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\int_{0}^{1} \frac{x ^{2}}{1 + x} d x

\int_{- \infty}^{0} (x + 1) e^{x} d x

\int_{- \infty}^{\frac{2}{a}} x e^{a x} d x

What is the integral from 0 to pi/4 of 3sec(x)tan(x) ?
The integral from 0 to pi/4 of 3sec(x)tan(x) is 3(sqrt(2)-1)