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Question Number 62185 by aliesam last updated on 17/Jun/19

$$\int \frac{dx}{sin3x+sin4x}$$

Answered by MJS last updated on 17/Jun/19

$$\int \frac{dx}{\sin 3x+\sin 4x}=$$$$[t=\frac{1}{\cos x}\to dx=\frac{{\cos }^{2}x}{\sin x}]$$$$=-\int \frac{t^3}{(t-1)(t+1)(t^3+4t^2-4t-8)}dt=$$$$\left[\begin{array}{c}\alpha =-\frac{4}{3}(1-\sqrt{7}\sin \left(\frac{1}{3}\arcsin \frac{\sqrt{7}}{14}\right))\\ \beta =\frac{4}{3}(-1+\sqrt{7}\cos (\frac{\pi }{6}+\arcsin \frac{\sqrt{7}}{14}))\\ \gamma =-\frac{4}{3}(1+\sqrt{7}\sin (\frac{\pi }{3}+\arcsin \frac{\sqrt{7}}{14}))\end{array}\right]$$$$=-\int \frac{t^3}{(t-1)(t+1)(t-\alpha )(t-\beta )(t-\gamma )}dt=$$$$=\frac{1}{14}\int \frac{dt}{t-1}+\frac{1}{2}\int \frac{dt}{t+1}+\frac{\gamma }{7}\int \frac{dt}{t-\alpha }+\frac{\alpha }{7}\int \frac{dt}{t-\beta }+\frac{\beta }{7}\int \frac{dt}{t-\gamma }=$$$$=\frac{1}{14}\ln (t-1)+\frac{1}{2}\ln (t+1)+\frac{\gamma }{7}\ln (t-\alpha )+\frac{\alpha }{7}\ln (t-\beta )+\frac{\beta }{7}\ln (t-\gamma )$$$$\mathrm{now}\mathrm{put}t=\frac{1}{\cos x}$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{solve}\mathrm{but}\mathrm{hard}\mathrm{to}\mathrm{write}\mathrm{out}[\mathrm{and}$$$$\mathrm{it}\mathrm{was}\mathrm{not}\mathrm{easy}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{constants}]$$

Commented by aliesam last updated on 17/Jun/19

$$thankyoysir$$$$godblessyou$$

Commented by MJS last updated on 17/Jun/19

$$\mathrm{I}\mathrm{tried}\mathrm{some}\mathrm{other}\mathrm{paths}\mathrm{but}\mathrm{none}\mathrm{worked}...$$

Commented by aliesam last updated on 17/Jun/19

$$thatoneisgoodsolutionthankyou$$

Commented by malwaan last updated on 18/Jun/19

$$GREATSIRMJS$$$$thankyousomuch$$

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