Use a table of integrals to find the indefinite integral ∫ex1−tan⁡exdx

Step 1 Given integral is ∫ex1−tan⁡exdx. Consider u=ex. u=ex du=exdx dx=duex Step 2 Now substitute the above values in the given integral. ∫ex1−tan⁡exdx=∫uduu(1−tan⁡(u)) =∫du(1−tan⁡(u)) Consider, v=tan⁡u2 dv=12sec2u2 Step 3 ∫ex1−tan⁡exdx=∫du(1−tan⁡(u)) =∫1−v2(1−v2−2v)(1+v2)dv =12ln|tan2(12ex)+2tan⁡(12ex)−1|+12ln⁡(sec2(12ex))+arctan⁡(tan⁡(12ex))+C

"Use a table of integrals to find the..." solved and explained

Ava-May Nelson

Answered question

2021-10-16

Use a table of integrals to find the indefinite integral

\int \frac{e^{x}}{1 - {\tan e}^{x}} d x

Answer & Explanation

wheezym

Skilled2021-10-17Added 103 answers

Step 1
Given integral is

\int \frac{e^{x}}{1 - {\tan e}^{x}} d x

.
Consider

u = e^{x}

u = e^{x}

d u = e^{x} d x

d x = \frac{d u}{e^{x}}

Step 2
Now substitute the above values in the given integral.

\int \frac{e^{x}}{1 - {\tan e}^{x}} d x = \int \frac{u d u}{u (1 - \tan (u))}

= \int \frac{d u}{(1 - \tan (u))}

Consider,

v = \tan \frac{u}{2}

d v = \frac{1}{2} \sec^{2} \frac{u}{2}

Step 3

\int \frac{e^{x}}{1 - {\tan e}^{x}} d x = \int \frac{d u}{(1 - \tan (u))}

= \int \frac{1 - v^{2}}{(1 - v^{2} - 2 v) (1 + v^{2})} d v

= \frac{1}{2} \ln | \tan^{2} (\frac{1}{2} e^{x}) + 2 \tan (\frac{1}{2} e^{x}) - 1 | + \frac{1}{2} \ln (\sec^{2} (\frac{1}{2} e^{x})) + \arctan (\tan (\frac{1}{2} e^{x})) + C

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