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Question Number 156319 by cortano last updated on 10/Oct/21

$$\int \frac{36}{{(12\cos \mathrm{x}+5\sin \mathrm{x})}^2}\mathrm{dx}=?$$

Commented by john_santu last updated on 10/Oct/21

$$=\frac{36}{13}\left(\frac{12\sin x-5\cos x}{5\sin x+12\cos x}\right)+c$$

Answered by puissant last updated on 10/Oct/21

$$=\int \frac{36}{{cos}^{2}x(12+5{tan}^{2}x)}dx\stackrel{u=tanx}{=}\int \frac{36}{12+5u^2}du$$$$=3\int \frac{du}{1+{\left(\sqrt{\frac{5}{12}}u\right)}^2}\stackrel{t=\sqrt{\frac{5}{12}}u}{=}\frac{3\sqrt{5}}{2\sqrt{3}}\int \frac{dt}{1+t^2}=\frac{\sqrt{15}}{2}arctan\left(t\right)+C$$$$$$$$\therefore \because K=\int \frac{36}{{(12cosx+5sinx)}^2}dx=\frac{\sqrt{15}}{2}arctan\left(\sqrt{\frac{5}{12}}tanx\right)+C$$

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