What is the integral from 0 to pi/4 of tan(2x+5) ?

The integral from 0 to pi/4 of tan(2x+5) is -1/2 (ln(cos(pi/2+5))-ln(cos(5)))

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integral from 0 to pi/4 of tan(2x+5)

\int_{0}^{\frac{π}{4}} tan (2 x + 5) d x

- \frac{1}{2} (ln (cos (\frac{π}{2} + 5)) - ln (cos (5)))

Solution steps

\int_{0}^{\frac{π}{4}} tan (2 x + 5) d x

Apply u-substitution

: \int_{5}^{\frac{π}{2} + 5} tan (u) \frac{1}{2} d u

= \int_{5}^{\frac{π}{2} + 5} tan (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{5}^{\frac{π}{2} + 5} tan (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int_{5}^{\frac{π}{2} + 5} \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int_{c o s (5)}^{c o s (\frac{π}{2} + 5)} - \frac{1}{v} d v

= \frac{1}{2} \cdot \int_{c o s (5)}^{c o s (\frac{π}{2} + 5)} - \frac{1}{v} d v

Take the constant out:

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

= \frac{1}{2} (- \int_{c o s (5)}^{c o s (\frac{π}{2} + 5)} \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} (- [ln ∣ v ∣]_{c o s (5)}^{c o s (\frac{π}{2} + 5)})

Simplify

= - \frac{1}{2} [ln ∣ v ∣]_{c o s (5)}^{c o s (\frac{π}{2} + 5)}

Compute the boundaries:

ln (cos (\frac{π}{2} + 5)) - ln (cos (5))

= - \frac{1}{2} (ln (cos (\frac{π}{2} + 5)) - ln (cos (5)))

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What is the integral from 0 to pi/4 of tan(2x+5) ?
The integral from 0 to pi/4 of tan(2x+5) is -1/2 (ln(cos(pi/2+5))-ln(cos(5)))