What is the integral of 1/(8sin^2(x)cos(x)) ?

The integral of 1/(8sin^2(x)cos(x)) is 1/8 (-csc(x)+ln|tan(x)+sec(x)|)+C

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integral of 1/(8sin^2(x)cos(x))

\int \frac{1}{8 sin ^{2} ( x ) cos ( x )} d x

\frac{1}{8} (- csc (x) + ln ∣ tan (x) + sec (x) ∣) + C

Solution steps

\int \frac{1}{8 sin ^{2} ( x ) cos ( x )} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{sin ^{2} ( x ) cos ( x )} d x

Rewrite using trig identities

= \frac{1}{8} \cdot \int \frac{csc ^{2} ( x )}{cos ( x )} d x

Apply Integration By Parts

: - csc (x) - \int - sec (x) d x

= \frac{1}{8} (- csc (x) - \int - sec (x) d x)

\int - sec (x) d x = - ln ∣ tan (x) + sec (x) ∣

= \frac{1}{8} (- csc (x) - (- ln ∣ tan (x) + sec (x) ∣))

Simplify

= \frac{1}{8} (- csc (x) + ln ∣ tan (x) + sec (x) ∣)

Add a constant to the solution

= \frac{1}{8} (- csc (x) + ln ∣ tan (x) + sec (x) ∣) + C

\int \frac{1}{x} d \frac{1}{x} d x

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\int \frac{1}{4 - 3 x ^{2}} d x

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\int \frac{y}{c - y} d y

What is the integral of 1/(8sin^2(x)cos(x)) ?
The integral of 1/(8sin^2(x)cos(x)) is 1/8 (-csc(x)+ln|tan(x)+sec(x)|)+C