How do you integrate $\int \frac{e^{x}}{7 + e^{2 x}}$ $int e^x/(7+e^(2x))$ by trigonometric substitution?

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How do you integrate $\int \frac{e^{x}}{7 + e^{2 x}}$ $int e^x/(7+e^(2x))$ by trigonometric substitution?

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Answer 1 · 2018-03-24T21:00:04

$\frac{1}{\sqrt{7}} arctan (\frac{e^{x}}{\sqrt{7}})$ $1/sqrt(7) arctan (e^ x/sqrt(7))$

Explanation:

$\int \frac{e^{x}}{7 + e^{2} x} d x = \int \frac{e^{x}}{7 + {(e^{x})}^{2}} d x$ $int e^x/(7+ e^2x)dx = int e^x/(7+ (e^x)^2)dx$

write it in a format that we can look to integrate more easily

Let $u = e^{x}$ $u = e^x$
$d u = e^{x} d x$ $du = e^x dx$

The integral in terms of u is now

$\int \frac{1}{7 + u^{2}} \cdot d u$ $int 1/(7+u^2)*du$

To use a trig substitution on this I used the fact that $\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} arctan (\frac{x}{a})$ $int 1/(a^2+x^2)dx = 1/aarctan(x/a)$

Therefore a good trig substitution for this sort of problem would be
to let $u = \sqrt{7} tan (A)$ $u = sqrt(7) tan(A)$ therefore $u^{2} = 7 {tan}^{2} (A)$ $u^2 = 7tan^2(A)$

$u = \sqrt{7} tan A$ $u = sqrt(7) tan A$

$u = \sqrt{7} \frac{sin (A)}{cos (A)}$ $u = sqrt(7) Sin(A)/Cos(A)$

differentiating using the quotient rule

$d u = \sqrt{7} (\frac{cos (A) cos (A) - sin (A) (- sin (A))}{{cos}^{2}}) d A$ $du = sqrt(7) ((cos(A)cos(A) - sin(A)(-sin(A)))/cos^2) dA$

$N o t e {cos}^{2} (A) + {sin}^{2} (A) = 1$ $Note cos^2(A) + sin ^2(A) = 1$

$d u = \sqrt{7} \frac{1}{{cos}^{2} (A)} d A$ $du = sqrt(7) 1/cos^2(A) dA$

The integral is now of the form

$\int \frac{1}{7} (1 + {tan}^{2} (A)) . \sqrt{7} \frac{1}{{cos}^{2} (A)} d A$ $int 1/7(1+tan^2(A)).sqrt(7) 1/cos^2(A) dA$

$= \int \frac{\sqrt{7}}{7} (1 + \frac{{sin}^{2} (A)}{{cos}^{2} (A)}) . \sqrt{7} \frac{1}{{cos}^{2} (A)} d A$ $= int sqrt(7)/7(1+sin^2(A)/cos^2(A)).sqrt(7) 1/cos^2(A) dA$

$= \int \frac{\sqrt{7}}{7} \cdot \frac{1}{\frac{{cos}^{2} (A) + {sin}^{2} (A)}{{cos}^{2} (A)}} \frac{1}{{cos}^{2} (A)} d A$ $= int sqrt(7)/7*1/((cos^2(A)+sin^2(A))/cos^2(A)) 1/cos^2(A) dA$

$= \int \frac{\sqrt{7}}{7} \cdot 1 d A$ $= int sqrt(7)/7*1 dA$ (everything cancels out)

$= \frac{\sqrt{7}}{7} A$ $= sqrt(7)/7A$

Now remember
$u = \sqrt{7} tan (A)$ $u = sqrt(7) tan(A)$ and $u^{2} = 7 {tan}^{2} (A)$ $u^2 = 7tan^2(A)$

Therefore $tan (A) = \frac{u}{\sqrt{7}}$ $tan(A) = u / sqrt(7)$

A = $arctan (\frac{u}{\sqrt{7}})$ $arctan (u/sqrt(7))$

So the integral results in = $\frac{\sqrt{7}}{7} arctan (\frac{u}{\sqrt{7}})$ $sqrt(7)/7arctan (u/sqrt(7))$

= $\frac{1}{\sqrt{7}} arctan (\frac{u}{\sqrt{7}})$ $1/sqrt(7)arctan (u/sqrt(7))$

finally we had $u = e^{x}$ $u = e^x$

Therefore the answer is $\frac{1}{\sqrt{7}} arctan (\frac{e^{x}}{\sqrt{7}})$ $1/sqrt(7) arctan (e^ x/sqrt(7))$

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