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# Find the integral

$\large \int \dfrac{ (\sin x + \cos x)^4}{(\cos x - \sin x)^4} \, dx = \, ?$

Note by Alaa Yousof
1 year ago

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For 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? · 1 year ago

no what are the next steps please · 1 year ago

What do you mean? · 1 year ago

how will we use the partial fractions · 1 year ago