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

The integral of \sqrt[7]{tan(5x)}sec^2(5x) is 7/40 tan^{8/7}(5x)+C

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

\frac{7}{40} tan^{\frac{8}{7}} (5 x) + C

Solution steps

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{7}{5} \cdot \frac{u ^{8}}{8}

Substitute back

Simplify

= \frac{7}{40} tan^{\frac{8}{7}} (5 x)

Add a constant to the solution

= \frac{7}{40} tan^{\frac{8}{7}} (5 x) + C

\int \frac{2}{( 2 x - 1 ) ^{2} + 4} d x

cos^{2} (x) sin (x)

\int \frac{- 8}{x x ^{2} - 36} d x

\int - 7 cos (x) d x

\int (2 x - 5)^{5} d x

What is the integral of \sqrt[7]{tan(5x)}sec^2(5x) ?
The integral of \sqrt[7]{tan(5x)}sec^2(5x) is 7/40 tan^{8/7}(5x)+C