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

The integral from 0 to pi/4 of tan(2θ) is diverges

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integral from 0 to pi/4 of tan(2θ)

\int_{0}^{\frac{π}{4}} tan (2 θ) d θ

diverges

Solution steps

\int_{0}^{\frac{π}{4}} tan (2 θ) d θ

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{sin ( 2 θ )}{cos ( 2 θ )} d θ

Apply u-substitution

: \int_{1}^{0} - \frac{1}{2 u} d u

= \int_{1}^{0} - \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{1}{2 u} 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_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{2} [ln ∣ u ∣]_{0}^{1})

Simplify

= \frac{1}{2} [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 θ) d θ

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

\int_{1}^{2} \frac{x}{( 3 x ^{2} - 1 ) ^{2}} d x

\int_{0}^{2} x (16 - x^{4})^{5} d x

\int_{- 2}^{0} (3 t^{3} - t^{2} - 9 t - 1) d t

What is the integral from 0 to pi/4 of tan(2θ) ?
The integral from 0 to pi/4 of tan(2θ) is diverges