What is the integral of cot^{-4}(x ?

The integral of cot^{-4}(x is (x^2tan^4(d))/2+C

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integral of cot^{-4}(x

\int cot^{- 4} (d) x d x

\frac{x ^{2} tan ^{4} ( d )}{2} + C

Solution steps

\int cot^{- 4} (d) x d x

Take the constant out:

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

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{cot ^{4} ( d )} \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= \frac{1}{cot ^{4} ( d )} \cdot \frac{x ^{2}}{2}

Simplify

\frac{1}{cot ^{4} ( d )} \cdot \frac{x ^{2}}{2} : \frac{x ^{2} tan ^{4} ( d )}{2}

= \frac{x ^{2} tan ^{4} ( d )}{2}

Add a constant to the solution

= \frac{x ^{2} tan ^{4} ( d )}{2} + C

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What is the integral of cot^{-4}(x ?
The integral of cot^{-4}(x is (x^2tan^4(d))/2+C