What is the integral from 0 to pi/(20) of (1+e^{tan(5x)})sec^2(5x) ?

The integral from 0 to pi/(20) of (1+e^{tan(5x)})sec^2(5x) is e/5

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integral from 0 to pi/(20) of (1+e^{tan(5x)})sec^2(5x)

\int_{0}^{\frac{π}{20}} (1 + e^{t a n (5 x)}) sec^{2} (5 x) d x

\frac{e}{5}

Decimal

0.54365 \dots

Solution steps

\int_{0}^{\frac{π}{20}} (1 + e^{t a n (5 x)}) sec^{2} (5 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{4}} (1 + e^{t a n (u)}) sec^{2} (u) \frac{1}{5} d u

= \int_{0}^{\frac{π}{4}} (1 + 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_{0}^{\frac{π}{4}} (1 + e^{t a n (u)}) sec^{2} (u) d u

Apply u-substitution

: \int_{0}^{1} 1 + e^{v} d v

= \frac{1}{5} \cdot \int_{0}^{1} 1 + e^{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{5} (\int_{0}^{1} 1 d v + \int_{0}^{1} e^{v} d v)

\int_{0}^{1} 1 d v = 1

\int_{0}^{1} e^{v} d v = e - 1

= \frac{1}{5} (1 + e - 1)

Simplify

\frac{1}{5} (1 + e - 1) : \frac{e}{5}

= \frac{e}{5}

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What is the integral from 0 to pi/(20) of (1+e^{tan(5x)})sec^2(5x) ?
The integral from 0 to pi/(20) of (1+e^{tan(5x)})sec^2(5x) is e/5