What is the integral of (cos^4(x))/(sin^8(x)) ?

The integral of (cos^4(x))/(sin^8(x)) is -1/7 cot^7(x)-1/5 cot^5(x)+C

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integral of (cos^4(x))/(sin^8(x))

\int \frac{cos ^{4} ( x )}{sin ^{8} ( x )} d x

- \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x) + C

Solution steps

\int \frac{cos ^{4} ( x )}{sin ^{8} ( x )} d x

Rewrite using trig identities

= \int \frac{cot ^{4} ( x )}{sin ^{4} ( x )} d x

Use the basic trigonometric identity:

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

= \int cot^{4} (x) csc^{4} (x) d x

Apply Trigonometric Substitution

: \int - u^{4} (u^{2} + 1) d u

= \int - u^{4} (u^{2} + 1) d u

Take the constant out:

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

= - \int u^{4} (u^{2} + 1) d u

Expand

u^{4} (u^{2} + 1) : u^{6} + u^{4}

= - \int u^{6} + u^{4} 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

= - (\int u^{6} d u + \int u^{4} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{4} d u = \frac{u ^{5}}{5}

= - (\frac{u ^{7}}{7} + \frac{u ^{5}}{5})

Substitute back

u = cot (x)

= - (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{5} ( x )}{5})

Simplify

- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{5} ( x )}{5}) : - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x)

= - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x)

Add a constant to the solution

= - \frac{1}{7} cot^{7} (x) - \frac{1}{5} cot^{5} (x) + C

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What is the integral of (cos^4(x))/(sin^8(x)) ?
The integral of (cos^4(x))/(sin^8(x)) is -1/7 cot^7(x)-1/5 cot^5(x)+C