What is the simplest way to compute : ∫0π(sin⁡(x)5−4cos⁡(x))2dx=π24

Let z=eix and then∫0π(sin⁡(x)5−4cos⁡(x))2dx=12∫{−π}π(sin⁡(x)5−4cos⁡(x))2dx=12∫|z|=1(z−1z2i5−4z+1z2)2dziz=12∫|z|=1(z2−12i(5z−2z2−2))2dziz=−18i∫|z|=1(z2−12z2−5z+2)2dzz=−18i∫|z|=1(z2−1)2z(2z−1)2(z−2)2dz=−132i∫|z|=1(z2−1)2z(z−12)2(z−2)2dz=−132i⋅2πi(Res((z2−1)2z(z−12)2(z−2)2,z=0)+Res((z2−1)2z(z−12)2(z−2)2,z=12))=−π16(1−53)=π24

Explained: "What is the simplest way to compute :..."

Asher Olsen

Answered question

2022-03-24

What is the simplest way to compute :

\int_{0}^{π} {(\frac{\sin (x)}{5 - 4 \cos (x)})}^{2} d x = \frac{π}{24}

Answer & Explanation

Cecilia Nolan

Beginner2022-03-25Added 13 answers

Let

z = e^{i x}

and then

\int_{0}^{π} {(\frac{\sin (x)}{5 - 4 \cos (x)})}^{2} d x

= \frac{1}{2} \int_{{- π}}^{π} {(\frac{\sin (x)}{5 - 4 \cos (x)})}^{2} d x

= \frac{1}{2} \int_{| z | = 1} {(\frac{\frac{z - \frac{1}{z}}{2 i}}{5 - 4 \frac{z + \frac{1}{z}}{2}})}^{2} \frac{d z}{i z}

= \frac{1}{2} \int_{| z | = 1} {(\frac{z^{2} - 1}{2 i (5 z - 2 z^{2} - 2)})}^{2} \frac{d z}{i z}

= - \frac{1}{8 i} \int_{| z | = 1} {(\frac{z^{2} - 1}{2 z^{2} - 5 z + 2})}^{2} \frac{d z}{z}

= - \frac{1}{8 i} \int_{| z | = 1} \frac{{(z^{2} - 1)}^{2}}{z {(2 z - 1)}^{2} {(z - 2)}^{2}} d z

= - \frac{1}{32 i} \int_{| z | = 1} \frac{{(z^{2} - 1)}^{2}}{z {(z - \frac{1}{2})}^{2} {(z - 2)}^{2}} d z

= - \frac{1}{32 i} \cdot 2 π i (Res (\frac{{(z^{2} - 1)}^{2}}{z {(z - \frac{1}{2})}^{2} {(z - 2)}^{2}}, z = 0) + Res (\frac{{(z^{2} - 1)}^{2}}{z {(z - \frac{1}{2})}^{2} {(z - 2)}^{2}}, z = \frac{1}{2}))

= - \frac{π}{16} (1 - \frac{5}{3})

= \frac{π}{24}

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