What is the integral of 1/(e^xcos^2(e^{-x))} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(e^xcos^2(e^{-x))}

\int \frac{1}{e ^{x} cos ^{2} ( e ^{- x} )} d x

- tan (e^{- x}) + C

Solution steps

\int \frac{1}{e ^{x} cos ^{2} ( e ^{- x} )} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{x} cos ^{2} ( e ^{- x} )} = e^{- x} \frac{1}{cos ^{2} ( e ^{- x} )}

= \int e^{- x} \frac{1}{cos ^{2} ( e ^{- x} )} d x

Simplify

e^{- x} \frac{1}{cos ^{2} ( e ^{- x} )} : e^{- x} sec^{2} (e^{- x})

= \int e^{- x} sec^{2} (e^{- x}) d x

Apply u-substitution

= \int - sec^{2} (u) d u

Take the constant out:

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

= - \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= - tan (u)

Substitute back

u = e^{- x}

= - tan (e^{- x})

Add a constant to the solution

= - tan (e^{- x}) + C

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