What is the integral of cot^7(x/4)csc^2(x/4) ?

The integral of cot^7(x/4)csc^2(x/4) is -1/2 cot^8(x/4)+C

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integral of cot^7(x/4)csc^2(x/4)

\int cot^{7} (\frac{x}{4}) csc^{2} (\frac{x}{4}) d x

- \frac{1}{2} cot^{8} (\frac{x}{4}) + C

Solution steps

\int cot^{7} (\frac{x}{4}) csc^{2} (\frac{x}{4}) d x

Apply u-substitution

: \int cot^{7} (u) csc^{2} (u) \cdot 4 d u

= \int cot^{7} (u) csc^{2} (u) \cdot 4 d u

Take the constant out:

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

= 4 \cdot \int cot^{7} (u) csc^{2} (u) d u

Apply Trigonometric Substitution

: \int - v^{7} d v

= 4 \cdot \int - v^{7} d v

Take the constant out:

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

= 4 (- \int v^{7} d v)

Apply the Power Rule

: \frac{v ^{8}}{8}

= 4 (- \frac{v ^{8}}{8})

Substitute back

= 4 (- \frac{cot ^{8} ( \frac{x}{4} )}{8})

Simplify

4 (- \frac{cot ^{8} ( \frac{x}{4} )}{8}) : - \frac{1}{2} cot^{8} (\frac{x}{4})

= - \frac{1}{2} cot^{8} (\frac{x}{4})

Add a constant to the solution

= - \frac{1}{2} cot^{8} (\frac{x}{4}) + C

\int x^{8} d x

\int \frac{3 x ^{2} - 7 x + 4}{x} d x

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

\int - 3 x e^{x^{2}} d x

\int \frac{7}{2} - x d x

What is the integral of cot^7(x/4)csc^2(x/4) ?
The integral of cot^7(x/4)csc^2(x/4) is -1/2 cot^8(x/4)+C