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

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

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

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

\frac{1}{64} (- csc (x) - sin (x)) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int 1 d u = u

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

Substitute back

u = sin (x)

= \frac{1}{64} (- \frac{1}{sin ( x )} - sin (x))

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= \frac{1}{64} (- csc (x) - sin (x))

Add a constant to the solution

= \frac{1}{64} (- csc (x) - sin (x)) + C

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