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

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

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

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

- \frac{1}{2} cot^{2} (7 x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{u}{7} d u

= 7 \cdot \int - \frac{u}{7} d u

Take the constant out:

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

= 7 (- \frac{1}{7} \cdot \int u d u)

Apply the Power Rule

: \frac{u ^{2}}{2}

= 7 (- \frac{1}{7} \cdot \frac{u ^{2}}{2})

Substitute back

u = cot (7 x)

= 7 (- \frac{1}{7} \cdot \frac{cot ^{2} ( 7 x )}{2})

Simplify

7 (- \frac{1}{7} \cdot \frac{cot ^{2} ( 7 x )}{2}) : - \frac{1}{2} cot^{2} (7 x)

= - \frac{1}{2} cot^{2} (7 x)

Add a constant to the solution

= - \frac{1}{2} cot^{2} (7 x) + C

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