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Question Number 39230 by MJS last updated on 04/Jul/18

$$\int \frac{d\alpha }{\sin 2\alpha +\sin 3\alpha +\sin 5\alpha }=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jul/18

$$sin2\alpha +sin3\alpha +sin5\alpha$$$$sin2\alpha +sin3\alpha +sin2\alpha cos3\alpha +cos2\alpha sin3\alpha$$$$sin2\alpha (1+cos3\alpha )+sin3\alpha (1+cos2\alpha )$$$$2sin\alpha cos\alpha .2{cos}^2\frac{3\alpha }{2}+2sin\frac{3\alpha }{2}cos\frac{3\alpha }{2}2{cos}^2\alpha$$$$4cos\alpha cos\frac{3\alpha }{2}(sin\alpha cos\frac{3\alpha }{2}+cos\alpha sin\frac{3\alpha }{2})$$$$4cos\alpha cos\frac{3\alpha }{2}sin\frac{5\alpha }{2}$$$$wait...$$$$$$

Answered by MJS last updated on 04/Jul/18

$$\sin 2\alpha =2\sin \alpha \cos \alpha$$$$\sin 3\alpha =3\sin \alpha {\cos }^2\alpha -{\sin }^3\alpha$$$$\sin 5\alpha ={\sin }^5\alpha -{10\sin }^3\alpha {\cos }^2\alpha +5\sin \alpha {\cos }^4\alpha$$$$\sin 2\alpha +\sin 3\alpha +\sin 5\alpha =$$$$2sc+3{sc}^2-s^3+s^5-10s^3{c}^2+5{sc}^4=$$$$=2sc+s(3c^2-s^2+s^4-10s^2{c}^2+5c^4)=$$$$=2sc+s(3c^2-(1-c^2)+{(1-c^2)}^2-10(1-c^2)c^2+5c^4)=$$$$=2sc+8{sc}^2(2c^2-1)=2\frac{s^{2}c+4s^2{c}^2(2c^2-1)}{s}=$$$$=2\frac{(1-c^2)c+4(1-c^2)c^2(2c^2-1)}{s}=$$$$=-2\frac{(4c^2+2c-1)(2c-1)(c-1)(c+1)c}{s}$$$$$$$$\int \frac{d\alpha }{\sin 2\alpha +\sin 3\alpha +\sin 5\alpha }=$$$$=-\frac{1}{2}\int \frac{\sin \alpha }{({4\cos }^2\alpha +2\cos \alpha -1)(2\cos \alpha -1)(\cos \alpha -1)(\cos \alpha +1)\cos \alpha }d\alpha =$$$$[t=\cos \alpha \to d\alpha =-\frac{dt}{\sin \alpha }]$$$$=\frac{1}{2}\int \frac{dt}{(4t^2+2t-1)(2t-1)(t-1)(t+1)t}=$$$$=\frac{1}{2}\int (\frac{C_1}{t}+\frac{C_2}{t-1}+\frac{C_3}{t+1}+\frac{C_4}{2t-1}+\frac{C_5}{2t+\frac{1-\sqrt{5}}{2}}+\frac{C_6}{2t+\frac{1+\sqrt{5}}{2}})dt=$$$$[C_1=-1;C_2=\frac{1}{10};C_3=-\frac{1}{6};C_4=-\frac{8}{3};C_5=\frac{12+4\sqrt{5}}{5};C_6=\frac{12-4\sqrt{5}}{5}]$$$$=-\frac{1}{2}\int \frac{dt}{t}+\frac{1}{20}\int \frac{dt}{t-1}-\frac{1}{12}\int \frac{dt}{t+1}-\frac{2}{3}\int \frac{dt}{t-\frac{1}{2}}+\frac{3+\sqrt{5}}{5}\int \frac{dt}{t+\frac{1-\sqrt{5}}{4}}+\frac{3-\sqrt{5}}{5}\int \frac{dt}{t+\frac{1+\sqrt{5}}{4}}=$$$$=-\frac{1}{2}\ln \mid t\mid +\frac{1}{20}\ln \mid t-1\mid -\frac{1}{12}\ln \mid t+1\mid -\frac{2}{3}\ln \mid t-\frac{1}{2}\mid +\frac{3+\sqrt{5}}{5}\ln \mid t+\frac{1-\sqrt{5}}{4}\mid +\frac{3-\sqrt{5}}{5}\ln \mid t+\frac{1+\sqrt{5}}{4}\mid =$$$$=-\frac{1}{2}\ln \mid \cos \alpha \mid +\frac{1}{20}\ln \mid \cos \alpha -1\mid -\frac{1}{12}\ln \mid \cos \alpha +1\mid -\frac{2}{3}\ln \mid \cos \alpha -\frac{1}{2}\mid +\frac{3+\sqrt{5}}{5}\ln \mid \cos \alpha +\frac{1-\sqrt{5}}{4}\mid +\frac{3-\sqrt{5}}{5}\ln \mid \cos \alpha +\frac{1+\sqrt{5}}{4}\mid +C$$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Jul/18

$$excellent...itsaproofofpatience...$$

Commented by MJS last updated on 04/Jul/18

$$\mathrm{corrected}$$

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