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

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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

= \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

\int y cos (y) d y

\int_{0}^{\infty} e^{- 0.5 x} d x

\int \frac{cos ( x )}{2 x} d x

\int e^{x} - x d x

2 x y d y + (x^{2} - y^{2}) d x = 0

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