What is the integral of (sec^2(1/(x^9)))/(x^{10)} ?

The integral of (sec^2(1/(x^9)))/(x^{10)} is -1/9 tan(1/(x^9))+C

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integral of (sec^2(1/(x^9)))/(x^{10)}

\int \frac{sec ^{2} ( \frac{1}{x ^{9}} )}{x ^{10}} d x

- \frac{1}{9} tan (\frac{1}{x ^{9}}) + C

Solution steps

\int \frac{sec ^{2} ( \frac{1}{x ^{9}} )}{x ^{10}} d x

Apply u-substitution

: \int \frac{sec ^{2} ( \frac{1}{u} )}{9 u ^{2}} d u

= \int \frac{sec ^{2} ( \frac{1}{u} )}{9 u ^{2}} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{sec ^{2} ( \frac{1}{u} )}{u ^{2}} d u

Apply u-substitution

: \int - sec^{2} (v) d v

= \frac{1}{9} \cdot \int - sec^{2} (v) d v

Take the constant out:

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

= \frac{1}{9} (- \int sec^{2} (v) d v)

Use the common integral:

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

= \frac{1}{9} (- tan (v))

Substitute back

= \frac{1}{9} (- tan (\frac{1}{x ^{9}}))

Simplify

= - \frac{1}{9} tan (\frac{1}{x ^{9}})

Add a constant to the solution

= - \frac{1}{9} tan (\frac{1}{x ^{9}}) + C

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What is the integral of (sec^2(1/(x^9)))/(x^{10)} ?
The integral of (sec^2(1/(x^9)))/(x^{10)} is -1/9 tan(1/(x^9))+C