What is the integral of e^{tan(5x)}sec^2(5x) ?

The integral of e^{tan(5x)}sec^2(5x) is 1/5 e^{tan(5x)}+C

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integral of e^{tan(5x)}sec^2(5x)

\int e^{t a n (5 x)} sec^{2} (5 x) d x

\frac{1}{5} e^{t a n (5 x)} + C

Solution steps

\int e^{t a n (5 x)} sec^{2} (5 x) d x

Apply u-substitution

: \int e^{t a n (u)} sec^{2} (u) \frac{1}{5} d u

= \int e^{t a n (u)} sec^{2} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int e^{t a n (u)} sec^{2} (u) d u

Apply u-substitution

: \int e^{v} d v

= \frac{1}{5} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{5} e^{v}

Substitute back

= \frac{1}{5} e^{t a n (5 x)}

Add a constant to the solution

= \frac{1}{5} e^{t a n (5 x)} + C

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

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

\frac{\partial}{\partial x} (8 x^{5} y^{6} + 6 x^{4} y^{8})

6 x^{- 4}

\frac{\partial}{\partial x} (5 y^{\frac{1}{4}} x^{\frac{1}{2}})

What is the integral of e^{tan(5x)}sec^2(5x) ?
The integral of e^{tan(5x)}sec^2(5x) is 1/5 e^{tan(5x)}+C