What is the integral of \sqrt[7]{tan(9x)}sec^2(9x) ?

The integral of \sqrt[7]{tan(9x)}sec^2(9x) is 7/72 tan^{8/7}(9x)+C

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integral of \sqrt[7]{tan(9x)}sec^2(9x)

\frac{7}{72} tan^{\frac{8}{7}} (9 x) + C

Solution steps

Apply u-substitution

: \int \frac{7 u ^{7}}{9} d u

= \int \frac{7 u ^{7}}{9} d u

Take the constant out:

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

= \frac{7}{9} \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{7}{9} \cdot \frac{u ^{8}}{8}

Substitute back

Simplify

= \frac{7}{72} tan^{\frac{8}{7}} (9 x)

Add a constant to the solution

= \frac{7}{72} tan^{\frac{8}{7}} (9 x) + C

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What is the integral of \sqrt[7]{tan(9x)}sec^2(9x) ?
The integral of \sqrt[7]{tan(9x)}sec^2(9x) is 7/72 tan^{8/7}(9x)+C