What is the integral of (7^{cot(2x)})/(sin^2(2x)) ?

The integral of (7^{cot(2x)})/(sin^2(2x)) is -(7^{cot(2x)})/(2ln(7))+C

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integral of (7^{cot(2x)})/(sin^2(2x))

\int \frac{7 ^{c o t (2 x)}}{sin ^{2} ( 2 x )} d x

- \frac{7 ^{c o t (2 x)}}{2 ln ( 7 )} + C

Solution steps

\int \frac{7 ^{c o t (2 x)}}{sin ^{2} ( 2 x )} d x

Apply u-substitution

: \int \frac{7 ^{c o t (u)}}{2 sin ^{2} ( u )} d u

= \int \frac{7 ^{c o t (u)}}{2 sin ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{7 ^{c o t (u)}}{sin ^{2} ( u )} d u

Use the basic trigonometric identity:

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

= \frac{1}{2} \cdot \int 7^{c o t (u)} csc^{2} (u) d u

Apply u-substitution

: \int - 7^{v} d v

= \frac{1}{2} \cdot \int - 7^{v} d v

Take the constant out:

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

= \frac{1}{2} (- \int 7^{v} d v)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{2} (- \frac{7 ^{v}}{ln ( 7 )})

Substitute back

= \frac{1}{2} (- \frac{7 ^{c o t (2 x)}}{ln ( 7 )})

Simplify

\frac{1}{2} (- \frac{7 ^{c o t (2 x)}}{ln ( 7 )}) : - \frac{7 ^{c o t (2 x)}}{2 ln ( 7 )}

= - \frac{7 ^{c o t (2 x)}}{2 ln ( 7 )}

Add a constant to the solution

= - \frac{7 ^{c o t (2 x)}}{2 ln ( 7 )} + C

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What is the integral of (7^{cot(2x)})/(sin^2(2x)) ?
The integral of (7^{cot(2x)})/(sin^2(2x)) is -(7^{cot(2x)})/(2ln(7))+C