What is the integral of (csc^2(x))/(cot^4(x)) ?

The integral of (csc^2(x))/(cot^4(x)) is (tan^3(x))/3+C

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integral of (csc^2(x))/(cot^4(x))

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

\frac{tan ^{3} ( x )}{3} + C

Solution steps

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

Rewrite using trig identities

= \int \frac{sin ^{4} ( x ) csc ^{2} ( x )}{cos ^{4} ( x )} d x

Simplify

\frac{sin ^{4} ( x ) csc ^{2} ( x )}{cos ^{4} ( x )} : sec^{4} (x) sin^{2} (x)

= \int sec^{4} (x) sin^{2} (x) d x

Rewrite using trig identities

= \int \frac{tan ^{2} ( x )}{cos ^{2} ( x )} d x

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = tan (x)

= \frac{tan ^{3} ( x )}{3}

Add a constant to the solution

= \frac{tan ^{3} ( x )}{3} + C

f (x) = 10 x^{3} - 20 x + \frac{1}{5}

y^{^{''}} + y = \frac{1}{4} sin (3 t) - \frac{3}{4} sin (t)

\frac{d}{d x} (arccos (\frac{a x}{2}))

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

\int_{0}^{\infty} cos^{2} (x) d x

What is the integral of (csc^2(x))/(cot^4(x)) ?
The integral of (csc^2(x))/(cot^4(x)) is (tan^3(x))/3+C