So we're integrating the final expression in the equation above. Use half angle tangent substitution, we have \( t = \tan(x) \Rightarrow dx = \dfrac{dt}{1+t^2} , \sin(2x) = \dfrac{2t}{1+t^2} \). Use partial fractions to finish it off, note that \(2t - 1 - t^2 = -(t-1)^2 \).

Can you finish it off from here?
–
Pi Han Goh
·
1 year, 3 months ago

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TopNewestFor simplicity sake, let \(s\) and \(c\) denote the functions \(\sin x , \cos x\) respectively, then we have \(s^2 + c^2 = 1 , 2cs = \sin(2x) \).

Then, \[ \begin{eqnarray} (s+c)^4 &=& s^4 + c^4 + 4cs(s^2 + c^2) + 6(cs)^2 = (s^2 + c^2)^2- 2(cs)^2 + 4cs + 6(cs)^2 = 1 + 4cs + 4(cs)^2 \\ (s-c)^4 &= & s^4 + c^4 - 4cs(s^2 + c^2) + 6(cs)^2 = (s^2 + c^2)^2- 2(cs)^2 - 4cs + 6(cs)^2 = 1 - 4cs + 4(cs)^2\end{eqnarray} \]

Taking their ratio gives: \[ \begin{eqnarray} \dfrac{1 + 4cs + 4(cs)^2}{1 - 4cs + 4(cs)^2 } &=& 1 + \dfrac{8sc}{1 - 4cs + 4(cs)^2} \\ &=&1 + \dfrac{4\sin(2x)}{1 - 2\sin(2x) + \sin^2(2x)} = 1 + 4 \cdot \dfrac{\sin (2x)}{(\sin(2x) - 1)^2} \\ &=& 1 + 4 \left [\dfrac1{\sin(2x) - 1}+ \dfrac1{(\sin(2x) - 1)^2} \right ] \end{eqnarray} \]

So we're integrating the final expression in the equation above. Use half angle tangent substitution, we have \( t = \tan(x) \Rightarrow dx = \dfrac{dt}{1+t^2} , \sin(2x) = \dfrac{2t}{1+t^2} \). Use partial fractions to finish it off, note that \(2t - 1 - t^2 = -(t-1)^2 \).

Can you finish it off from here? – Pi Han Goh · 1 year, 3 months ago

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– Who Ting · 1 year, 3 months ago

no what are the next steps pleaseLog in to reply

– Pi Han Goh · 1 year, 3 months ago

What do you mean?Log in to reply

– Who Ting · 1 year, 3 months ago

how will we use the partial fractionsLog in to reply

– Pi Han Goh · 1 year, 3 months ago

I don't know where exactly you're stuck on. Please show your steps.Log in to reply