What is the integral of e^{tan(16x)}sec^2(16x) ?

The integral of e^{tan(16x)}sec^2(16x) is 1/16 e^{tan(16x)}+C

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integral of e^{tan(16x)}sec^2(16x)

\int e^{t a n (16 x)} sec^{2} (16 x) d x

\frac{1}{16} e^{t a n (16 x)} + C

Solution steps

\int e^{t a n (16 x)} sec^{2} (16 x) d x

Apply u-substitution

: \int e^{t a n (u)} sec^{2} (u) \frac{1}{16} d u

= \int e^{t a n (u)} sec^{2} (u) \frac{1}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{t a n (u)} sec^{2} (u) d u

Apply u-substitution

: \int e^{v} d v

= \frac{1}{16} \cdot \int e^{v} d v

Use the common integral:

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

= \frac{1}{16} e^{v}

Substitute back

= \frac{1}{16} e^{t a n (16 x)}

Add a constant to the solution

= \frac{1}{16} e^{t a n (16 x)} + C

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What is the integral of e^{tan(16x)}sec^2(16x) ?
The integral of e^{tan(16x)}sec^2(16x) is 1/16 e^{tan(16x)}+C