What is the integral of (cot(x))/(sin^7(x)+1) ?

The integral of (cot(x))/(sin^7(x)+1) is 1/7 (ln|sin^7(x)|-ln|1+sin^7(x)|)+C

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integral of (cot(x))/(sin^7(x)+1)

\int \frac{cot ( x )}{sin ^{7} ( x ) + 1} d x

\frac{1}{7} (ln sin^{7} (x) - ln 1 + sin^{7} (x)) + C

Solution steps

\int \frac{cot ( x )}{sin ^{7} ( x ) + 1} d x

Apply u-substitution

: \int \frac{1}{u ( u ^{7} + 1 )} d u

= \int \frac{1}{u ( u ^{7} + 1 )} d u

Apply u-substitution

: \int \frac{1}{7 ( v - 1 ) v} d v

= \int \frac{1}{7 ( v - 1 ) v} d v

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{( v - 1 ) v} d v

Take the partial fraction of

\frac{1}{( v - 1 ) v} : \frac{1}{v - 1} - \frac{1}{v}

= \frac{1}{7} \cdot \int \frac{1}{v - 1} - \frac{1}{v} d v

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}{7} (\int \frac{1}{v - 1} d v - \int \frac{1}{v} d v)

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{7} (ln ∣ v - 1 ∣ - ln ∣ v ∣)

Substitute back

= \frac{1}{7} (ln 1 + sin^{7} (x) - 1 - ln 1 + sin^{7} (x))

Simplify

\frac{1}{7} (ln 1 + sin^{7} (x) - 1 - ln 1 + sin^{7} (x)) : \frac{1}{7} (ln sin^{7} (x) - ln 1 + sin^{7} (x))

= \frac{1}{7} (ln sin^{7} (x) - ln 1 + sin^{7} (x))

Add a constant to the solution

= \frac{1}{7} (ln sin^{7} (x) - ln 1 + sin^{7} (x)) + C

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What is the integral of (cot(x))/(sin^7(x)+1) ?
The integral of (cot(x))/(sin^7(x)+1) is 1/7 (ln|sin^7(x)|-ln|1+sin^7(x)|)+C