Question #98482

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Question #98482

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Answer 1 · 2017-06-05T00:33:34

$\int \frac{{sin}^{2} (\frac{x}{2}) + {cos}^{2} (\frac{x}{2})}{sin (\frac{x}{2}) - cos (\frac{x}{2})} d x = \sqrt{2} ln ∣ ∣ ∣ ∣ \frac{tan (\frac{x}{4}) + 1 - \sqrt{2}}{tan (\frac{x}{4}) + 1 + \sqrt{2}} ∣ ∣ ∣ ∣ + C$ $int (sin^2(x/2)+cos^2(x/2))/(sin(x/2) - cos(x/2))dx = sqrt2 ln abs ((tan(x/4) +1 -sqrt(2))/(tan(x/4) +1 +sqrt2)) + C$

Explanation:

Based on the fundamental trigonometric identity:

${sin}^{2} α + {cos}^{2} α = 1$ $sin^2 alpha + cos^2 alpha = 1$

we have that:

$\int \frac{{sin}^{2} (\frac{x}{2}) + {cos}^{2} (\frac{x}{2})}{sin (\frac{x}{2}) - cos (\frac{x}{2})} d x = \int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})}$ $int (sin^2(x/2)+cos^2(x/2))/(sin(x/2) - cos(x/2))dx = int dx/(sin(x/2)- cos(x/2))$

Use now the parametric fomulas:

$sin (\frac{x}{2}) = \frac{2 tan (\frac{x}{4})}{1 + {tan}^{2} (\frac{x}{4})}$ $sin(x/2) = (2tan(x/4))/(1+tan^2(x/4))$

$cos (\frac{x}{2}) = \frac{1 - {tan}^{2} (\frac{x}{4})}{1 + {tan}^{2} (\frac{x}{4})}$ $cos(x/2) = (1-tan^2(x/4))/(1+tan^2(x/4))$

substituting:

$t = tan (\frac{x}{4})$ $t= tan(x/4)$

$x = 4 arctan (t)$ $x = 4arctan(t)$

$d x = \frac{4 d t}{1 + t^{2}}$ $dx = (4dt)/(1+t^2)$

we have:

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = \int \frac{1}{\frac{2 t}{1 + t^{2}} - \frac{1 - t^{2}}{1 + t^{2}}} \frac{4 d t}{1 + t^{2}}$ $int dx/(sin(x/2)- cos(x/2)) = int 1/( (2t)/(1+t^2) - (1-t^2)/(1+t^2)) (4dt)/(1+t^2)$

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = 4 \int \frac{d t}{t^{2} + 2 t - 1}$ $int dx/(sin(x/2)- cos(x/2)) = 4 int (dt)/(t^2+2t-1)$

Complete the square at the denominator:

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = 4 \int \frac{d t}{{(t + 1)}^{2} - 2} = 2 \int \frac{d t}{{(\frac{t + 1}{\sqrt{2}})}^{2} - 1}$ $int dx/(sin(x/2)- cos(x/2)) = 4 int (dt)/((t+1)^2-2) = 2int (dt)/(((t+1)/sqrt2)^2 -1)$

Substitute again:

$u = \frac{t + 1}{\sqrt{2}}$ $u = (t+1)/sqrt2$

$d t = \sqrt{2} d u$ $dt =sqrt2 du$

so:

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = 2 \sqrt{2} \int \frac{d u}{u^{2} - 1}$ $int dx/(sin(x/2)- cos(x/2)) = 2sqrt2 int (du)/(u^2-1)$

factorize the denominator and perform partial fractions decompositions:

$\frac{1}{u^{2} - 1} = \frac{1}{(u - 1) (u + 1)} = \frac{1}{2} (\frac{1}{u - 1} - \frac{1}{u + 1})$ $1/(u^2-1) = 1/((u-1)(u+1)) = 1/2(1/(u-1) - 1/(u+1))$

Then:

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = \sqrt{2} (\int \frac{d u}{u - 1} - \int \frac{d u}{u + 1})$ $int dx/(sin(x/2)- cos(x/2)) = sqrt2 (int (du)/(u-1) -int (du)/(u+1))$

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = \sqrt{2} (ln | u - 1 | - ln | u + 1 |) + C$ $int dx/(sin(x/2)- cos(x/2)) = sqrt2(ln abs (u-1) - ln abs (u+1)) +C$

using the properties of logarithms:

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = \sqrt{2} ln ∣ ∣ ∣ \frac{u - 1}{u + 1} ∣ ∣ ∣ + C$ $int dx/(sin(x/2)- cos(x/2)) = sqrt2 ln abs ((u-1)/(u+1))+C$

undoing the substitutions:

$u = \frac{tan (\frac{x}{4}) + 1}{\sqrt{2}}$ $u = (tan(x/4)+1)/sqrt2$

$\frac{u - 1}{u + 1} = \frac{\frac{tan (\frac{x}{4}) + 1}{\sqrt{2}} - 1}{\frac{tan (\frac{x}{4}) + 1}{\sqrt{2}} + 1} = \frac{tan (\frac{x}{4}) + 1 - \sqrt{2}}{tan (\frac{x}{4}) + 1 + \sqrt{2}}$ $(u-1)/(u+1) = ( ( tan(x/4)+1)/sqrt2 -1)/((tan(x/4)+1)/sqrt2+1) = (tan(x/4) +1 -sqrt(2))/(tan(x/4) +1 +sqrt2)$

Finally:

$\int \frac{d x}{sin (\frac{x}{2}) - cos (\frac{x}{2})} = \sqrt{2} ln ∣ ∣ ∣ ∣ \frac{tan (\frac{x}{4}) + 1 - \sqrt{2}}{tan (\frac{x}{4}) + 1 + \sqrt{2}} ∣ ∣ ∣ ∣ + C$ $int dx/(sin(x/2)- cos(x/2)) = sqrt2 ln abs ((tan(x/4) +1-sqrt(2))/(tan(x/4) +1 +sqrt2)) + C$

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