What is the integral of (c_{1})/(xcos^2(ln(x))) ?

The integral of (c_{1})/(xcos^2(ln(x))) is c_{1}tan(ln(x))+C

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integral of (c_{1})/(xcos^2(ln(x)))

\int \frac{c _{1}}{x cos ^{2} ( ln ( x ) )} d x

c_{1} tan (ln (x)) + C

Solution steps

\int \frac{c _{1}}{x cos ^{2} ( ln ( x ) )} d x

Take the constant out:

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

= c_{1} \cdot \int \frac{1}{x cos ^{2} ( ln ( x ) )} d x

Apply u-substitution

: \int sec^{2} (u) d u

= c_{1} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= c_{1} tan (u)

Substitute back

u = ln (x)

= c_{1} tan (ln (x))

Add a constant to the solution

= c_{1} tan (ln (x)) + C

x \to 1 lim (- x + 1)

\frac{d}{d x} (\frac{a e ^{\frac{x}{2}}}{2} - b e^{- x})

\frac{d}{d x} (\frac{1}{- 25 x ^{2} + 2 y ^{2} - 50})

x \to 4 lim (x + x)

y = ln (x x^{2} - 2)

What is the integral of (c_{1})/(xcos^2(ln(x))) ?
The integral of (c_{1})/(xcos^2(ln(x))) is c_{1}tan(ln(x))+C