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Question Number 68967 by aliesam last updated on 17/Sep/19

Answered by mind is power last updated on 17/Sep/19

$$tg\left(\frac{x}{2}\right)=t$$$$dx=\frac{2dt}{1+t^2}$$$$\Rightarrow sin\left(x\right)=\frac{2t}{1+t^2}cos\left(x\right)=\frac{1-t^2}{1+t^2}$$$$\Rightarrow \int \frac{dx}{5sin\left(x\right)-4cos\left(x\right)+3}=\int \frac{2dt}{3(1+t^2)+10t-4+4t^2}$$$$=\int \frac{dt}{7t^2+10t-1}=\int \frac{dt}{7(t+5-\sqrt{32})(t+5+\sqrt{32})}$$$$=\int [\frac{1}{14\sqrt{32}(t+5-\sqrt{32})}-\frac{1}{14\sqrt{32}(t+5+\sqrt{32)}}]dt$$$$=\frac{1}{14\sqrt{32}}ln\mid \frac{t+5-\sqrt{32}}{t+5+\sqrt{32}}\mid +c$$$$$$$$\Rightarrow \int \frac{dx}{5sin\left(x\right)-4cos\left(x\right)+3}=\frac{1}{14\sqrt{32}}ln\mid \frac{tan\left(\frac{x}{2}\right)+5-\sqrt{32}}{tg\left(\frac{x}{2}\right)+5+\sqrt{32}}\mid +c$$$$$$

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