What is the integral of 1/(64sec^2(x)tan(x)) ?

The integral of 1/(64sec^2(x)tan(x)) is 1/64 (ln|sin(x)|-(sin^2(x))/2)+C

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integral of 1/(64sec^2(x)tan(x))

\int \frac{1}{64 sec ^{2} ( x ) tan ( x )} d x

\frac{1}{64} (ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2}) + C

Solution steps

\int \frac{1}{64 sec ^{2} ( x ) tan ( x )} d x

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{1}{sec ^{2} ( x ) tan ( x )} d x

Rewrite using trig identities

= \frac{1}{64} \cdot \int \frac{( 1 - sin ^{2} ( x ) ) cos ( x )}{sin ( x )} d x

Apply u-substitution

= \frac{1}{64} \cdot \int \frac{1 - u ^{2}}{u} d u

Expand

\frac{1 - u ^{2}}{u} : \frac{1}{u} - u

= \frac{1}{64} \cdot \int \frac{1}{u} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{64} (\int \frac{1}{u} d u - \int u d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= \frac{1}{64} (ln ∣ u ∣ - \frac{u ^{2}}{2})

Substitute back

u = sin (x)

= \frac{1}{64} (ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2})

Add a constant to the solution

= \frac{1}{64} (ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2}) + C

What is the integral of 1/(64sec^2(x)tan(x)) ?
The integral of 1/(64sec^2(x)tan(x)) is 1/64 (ln|sin(x)|-(sin^2(x))/2)+C