Step by step solution to "Evaluate the following 1.∫ex1−e2xdx 2. ∫x25+x2dx"

Harlen Pritchard

Answered question

2021-08-18

Evaluate the following
1.

\int e^{x} \sqrt{1 - e^{2 x}} d x

\int \frac{x^{2}}{\sqrt{5} + x^{2}} d x

pattererX

Skilled2021-08-19Added 95 answers

1.Finding the integral

\int e^{x} \sqrt{1 - e^{2 x} d x .}

Let us first apply substitution

u = e^{x}

we get

\int e^{x} \sqrt{1 - e^{2 x}} d x = \int \sqrt{1 - u^{2}} d u

Further apply the substitution

u = \sin v

we get

\int \sqrt{1 - u^{2}} d u = \int \cos^{2} (v) d v

\int \frac{1 + \cos (2 v)}{2} d v [: ‘ \cos^{2} (x) = \frac{1 + \cos (2 x)}{2}]

\frac{1}{2} \cdot (v + \frac{1}{2} \sin (2 v))

Substitute back

v = \arcsin (u)

, and

u = e^{x}

, we get

v = \arcsin (e^{x})

,we get

\int e^{x} \sqrt{1 - e^{2 x}} d x = \frac{1}{2} (\arcsin (e^{x}) + \frac{1}{2} \sin (2 \arcsin (e^{x}))) + C

2. Finding the integral

\int \frac{x^{2}}{\sqrt{5} + x^{2}} d x

\int \frac{x^{2}}{\sqrt{5} + x^{2}} d x = 5 \cdot \int \tan^{2} (u) \sec (u) d u [x = \sqrt{5} \tan (u)]

= 5 \cdot \int (- 1 + \sec^{2} (u)) \sec (u) d u [\tan^{2} (x) = - 1 + \sec^{2} (x)]

5 (\ln | \tan (u) + \sec (u) | + \frac{1}{2} \sec (u) \tan (u) + \frac{1}{2} \ln ∣ \tan (u) + \sec)

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