What is the integral of 1/(sin(5x)cos(5x)) ?

The integral of 1/(sin(5x)cos(5x)) is 1/5 ln|tan(5x)|+C

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integral of 1/(sin(5x)cos(5x))

\int \frac{1}{sin ( 5 x ) cos ( 5 x )} d x

\frac{1}{5} ln ∣ tan (5 x) ∣ + C

Solution steps

\int \frac{1}{sin ( 5 x ) cos ( 5 x )} d x

Rewrite using trig identities

= \int \frac{2}{sin ( 10 x )} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{sin ( 10 x )} d x

Rewrite using trig identities

= 2 \cdot \int csc (10 x) d x

Apply u-substitution

: \int csc (u) \frac{1}{10} d u

= 2 \cdot \int csc (u) \frac{1}{10} d u

Take the constant out:

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

= 2 \cdot \frac{1}{10} \cdot \int csc (u) d u

Apply u-substitution

: \int \frac{1}{v} d v

= 2 \cdot \frac{1}{10} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{10} ln ∣ v ∣

Substitute back

= 2 \cdot \frac{1}{10} ln tan (\frac{10 x}{2})

Simplify

2 \cdot \frac{1}{10} ln tan (\frac{10 x}{2}) : \frac{1}{5} ln ∣ tan (5 x) ∣

= \frac{1}{5} ln ∣ tan (5 x) ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ tan (5 x) ∣ + C

\int 8 e^{- t} d t

- \frac{x ^{2}}{20} + 230 x - 74000

y = x^{2} + 4

y^{^{''}} + 16 y = e^{3 x}

\int \frac{- sin ( 8 t )}{8} d t

What is the integral of 1/(sin(5x)cos(5x)) ?
The integral of 1/(sin(5x)cos(5x)) is 1/5 ln|tan(5x)|+C