What is the integral of (cos^3(x))/(sin^5(x)) ?

The integral of (cos^3(x))/(sin^5(x)) is -(cot^4(x))/4+C

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integral of (cos^3(x))/(sin^5(x))

\int \frac{cos ^{3} ( x )}{sin ^{5} ( x )} d x

- \frac{cot ^{4} ( x )}{4} + C

Solution steps

\int \frac{cos ^{3} ( x )}{sin ^{5} ( x )} d x

Rewrite using trig identities

= \int \frac{cot ^{3} ( x )}{sin ^{2} ( x )} d x

Use the basic trigonometric identity:

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

= \int cot^{3} (x) csc^{2} (x) d x

Apply u-substitution

: \int - u^{3} d u

= \int - u^{3} d u

Take the constant out:

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

= - \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - \frac{u ^{4}}{4}

Substitute back

u = cot (x)

= - \frac{cot ^{4} ( x )}{4}

Add a constant to the solution

= - \frac{cot ^{4} ( x )}{4} + C

\int \frac{( 2 + x ) ^{6}}{x} d x

\int \frac{( 2 - 4 x ) ^{2}}{x ^{2}} d x

\int \frac{1}{2 + 4 x ^{4}} (16 x^{3}) d x

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

\int \frac{x ^{2}}{( x ^{2} - 16 ) ^{\frac{5}{2}}} d x

What is the integral of (cos^3(x))/(sin^5(x)) ?
The integral of (cos^3(x))/(sin^5(x)) is -(cot^4(x))/4+C