What is the integral of \sqrt[5]{tan(7x)}sec^2(7x) ?

The integral of \sqrt[5]{tan(7x)}sec^2(7x) is 5/42 tan^{6/5}(7x)+C

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integral of \sqrt[5]{tan(7x)}sec^2(7x)

\int 5 tan (7 x) sec^{2} (7 x) d x

\frac{5}{42} tan^{\frac{6}{5}} (7 x) + C

Solution steps

\int 5 tan (7 x) sec^{2} (7 x) d x

Apply u-substitution

: \int \frac{5 u ^{5}}{7} d u

= \int \frac{5 u ^{5}}{7} d u

Take the constant out:

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

= \frac{5}{7} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{5}{7} \cdot \frac{u ^{6}}{6}

Substitute back

u = 5 tan (7 x)

= \frac{5}{7} \cdot \frac{( 5 tan ( 7 x ) ) ^{6}}{6}

Simplify

\frac{5}{7} \cdot \frac{( 5 tan ( 7 x ) ) ^{6}}{6} : \frac{5}{42} tan^{\frac{6}{5}} (7 x)

= \frac{5}{42} tan^{\frac{6}{5}} (7 x)

Add a constant to the solution

= \frac{5}{42} tan^{\frac{6}{5}} (7 x) + C

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What is the integral of \sqrt[5]{tan(7x)}sec^2(7x) ?
The integral of \sqrt[5]{tan(7x)}sec^2(7x) is 5/42 tan^{6/5}(7x)+C