What is the integral of 35csc(5x)sec(5x) ?

The integral of 35csc(5x)sec(5x) is 7/2 ln(tan^2(5x))+C

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integral of 35csc(5x)sec(5x)

\int 35 csc (5 x) sec (5 x) d x

\frac{7}{2} ln (tan^{2} (5 x)) + C

Solution steps

\int 35 csc (5 x) sec (5 x) d x

Take the constant out:

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

= 35 \cdot \int csc (5 x) sec (5 x) d x

Apply u-substitution

: \int csc (u) sec (u) \frac{1}{5} d u

= 35 \cdot \int csc (u) sec (u) \frac{1}{5} d u

Take the constant out:

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

= 35 \cdot \frac{1}{5} \cdot \int csc (u) sec (u) d u

Apply u-substitution

: \int \frac{v}{v ^{2} - 1} d v

= 35 \cdot \frac{1}{5} \cdot \int \frac{v}{v ^{2} - 1} d v

Apply u-substitution

: \int \frac{1}{2 w} d w

= 35 \cdot \frac{1}{5} \cdot \int \frac{1}{2 w} d w

Take the constant out:

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

= 35 \cdot \frac{1}{5} \cdot \frac{1}{2} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= 35 \cdot \frac{1}{5} \cdot \frac{1}{2} ln ∣ w ∣

Substitute back

= 35 \cdot \frac{1}{5} \cdot \frac{1}{2} ln sec^{2} (5 x) - 1

Simplify

35 \cdot \frac{1}{5} \cdot \frac{1}{2} ln sec^{2} (5 x) - 1 : \frac{7}{2} ln (tan^{2} (5 x))

= \frac{7}{2} ln (tan^{2} (5 x))

Add a constant to the solution

= \frac{7}{2} ln (tan^{2} (5 x)) + C

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What is the integral of 35csc(5x)sec(5x) ?
The integral of 35csc(5x)sec(5x) is 7/2 ln(tan^2(5x))+C