Find the integral

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

#Calculus

Note by Alaa Yousof
5 years, 3 months ago

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Comments

For simplicity sake, let ss and cc denote the functions sin⁡x,cos⁡x\sin x , \cos x respectively, then we have s2+c2=1,2cs=sin⁡(2x)s^2 + c^2 = 1 , 2cs = \sin(2x) .

Then, (s+c)4=s4+c4+4cs(s2+c2)+6(cs)2=(s2+c2)2−2(cs)2+4cs+6(cs)2=1+4cs+4(cs)2(s−c)4=s4+c4−4cs(s2+c2)+6(cs)2=(s2+c2)2−2(cs)2−4cs+6(cs)2=1−4cs+4(cs)2 \begin{aligned} (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{aligned}

Taking their ratio gives: 1+4cs+4(cs)21−4cs+4(cs)2=1+8sc1−4cs+4(cs)2=1+4sin⁡(2x)1−2sin⁡(2x)+sin⁡2(2x)=1+4⋅sin⁡(2x)(sin⁡(2x)−1)2=1+4[1sin⁡(2x)−1+1(sin⁡(2x)−1)2] \begin{aligned} \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{aligned}

So we're integrating the final expression in the equation above. Use half angle tangent substitution, we have t=tan⁡(x)⇒dx=dt1+t2,sin⁡(2x)=2t1+t2 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−t2=−(t−1)22t - 1 - t^2 = -(t-1)^2 .

Can you finish it off from here?

Pi Han Goh - 5 years, 3 months ago

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no what are the next steps please

who ting - 5 years, 2 months ago

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What do you mean?

Pi Han Goh - 5 years, 2 months ago

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@Pi Han Goh – how will we use the partial fractions

who ting - 5 years, 2 months ago

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@Who Ting – I don't know where exactly you're stuck on. Please show your steps.

Pi Han Goh - 5 years, 2 months ago
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