What is the integral of tan(x)[ln(cos(x))] ?

The integral of tan(x)[ln(cos(x))] is -(ln^2(cos(x)))/2+C

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integral of tan(x)[ln(cos(x))]

\int tan (x) [ln (cos (x))] d x

- \frac{ln ^{2} ( cos ( x ) )}{2} + C

Solution steps

\int tan (x) [ln (cos (x))] d x

Simplify

= \int tan (x) ln (cos (x)) d x

Apply u-substitution

: \int - \frac{ln ( u )}{u} d u

= \int - \frac{ln ( u )}{u} d u

Take the constant out:

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

= - \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= - \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= - \frac{v ^{2}}{2}

Substitute back

= - \frac{ln ^{2} ( cos ( x ) )}{2}

Add a constant to the solution

= - \frac{ln ^{2} ( cos ( x ) )}{2} + C

f (x) = cos (θ^{2})

x \to 5 lim (\frac{x ^{2} - 7 x + 10}{x ^{2} - 25})

\frac{\partial}{\partial x} (\frac{1}{4} x^{2} + y^{2})

y = 1 x

\int_{3}^{\infty} e^{- 4 x} d x

What is the integral of tan(x)[ln(cos(x))] ?
The integral of tan(x)[ln(cos(x))] is -(ln^2(cos(x)))/2+C