What is the integral of (4^{cot(5x)})/(sin^2(5x)) ?

The integral of (4^{cot(5x)})/(sin^2(5x)) is -(2^{2cot(5x)-1})/(5ln(2))+C

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integral of (4^{cot(5x)})/(sin^2(5x))

\int \frac{4 ^{c o t (5 x)}}{sin ^{2} ( 5 x )} d x

- \frac{2 ^{2 c o t (5 x) - 1}}{5 ln ( 2 )} + C

Solution steps

\int \frac{4 ^{c o t (5 x)}}{sin ^{2} ( 5 x )} d x

Apply u-substitution

: \int \frac{4 ^{c o t (u)}}{5 sin ^{2} ( u )} d u

= \int \frac{4 ^{c o t (u)}}{5 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}{5} \cdot \int \frac{4 ^{c o t (u)}}{sin ^{2} ( u )} d u

Use the basic trigonometric identity:

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

= \frac{1}{5} \cdot \int 4^{c o t (u)} csc^{2} (u) d u

Apply u-substitution

: \int - 4^{v} d v

= \frac{1}{5} \cdot \int - 4^{v} d v

Take the constant out:

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

= \frac{1}{5} (- \int 4^{v} d v)

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

= \frac{1}{5} (- \frac{4 ^{v}}{ln ( 4 )})

Substitute back

= \frac{1}{5} (- \frac{4 ^{c o t (5 x)}}{ln ( 4 )})

Simplify

\frac{1}{5} (- \frac{4 ^{c o t (5 x)}}{ln ( 4 )}) : - \frac{2 ^{2 c o t (5 x) - 1}}{5 ln ( 2 )}

= - \frac{2 ^{2 c o t (5 x) - 1}}{5 ln ( 2 )}

Add a constant to the solution

= - \frac{2 ^{2 c o t (5 x) - 1}}{5 ln ( 2 )} + C

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\int x^{2} - 9 x d x

\int \frac{1 - cos ( x )}{sin ( x )} d x

f (x) = (x - 6) (x + 9) (x - 8)

\int \frac{x - 9}{( x + 5 ) \cdot ( x - 2 )} d x

What is the integral of (4^{cot(5x)})/(sin^2(5x)) ?
The integral of (4^{cot(5x)})/(sin^2(5x)) is -(2^{2cot(5x)-1})/(5ln(2))+C