What is the integral from 0 to pi/4 of (1+e^{tan(x)})sec^2(x) ?

The integral from 0 to pi/4 of (1+e^{tan(x)})sec^2(x) is e

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integral from 0 to pi/4 of (1+e^{tan(x)})sec^2(x)

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

e

Decimal Notation

2.71828 \dots

Solution steps

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

Apply u-substitution

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

= \int_{0}^{1} 1 + e^{u} d u

Apply the Sum Rule:

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

= \int_{0}^{1} 1 d u + \int_{0}^{1} e^{u} d u

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

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

= 1 + e - 1

Simplify

= e

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What is the integral from 0 to pi/4 of (1+e^{tan(x)})sec^2(x) ?
The integral from 0 to pi/4 of (1+e^{tan(x)})sec^2(x) is e