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Question Number 105862 by bemath last updated on 01/Aug/20

$$\int \frac{dx}{9+16\cos {}^{2}x}?$$

Commented by bemath last updated on 01/Aug/20

$$thankyou$$

Answered by Mr.D.N. last updated on 01/Aug/20

$$\int \frac{\mathrm{dx}}{9+{16\cos }^2\mathrm{x}}=\int \frac{\frac{1}{{\cos }^2\mathrm{x}}}{\frac{9+{16\cos }^2\mathrm{x}}{{\cos }^2\mathrm{x}}}\mathrm{dx}$$$$\int \frac{{\sec }^2\mathrm{x}\mathrm{dx}}{{9\sec }^2\mathrm{x}+16}=\int \frac{{\sec }^2\mathrm{xdx}}{9(1+{\tan }^2\mathrm{x})+16}$$$$=\int \frac{{\sec }^2\mathrm{xdx}}{{9\tan }^2\mathrm{x}+25}=\frac{1}{9}\int \frac{{\sec }^2\mathrm{xdx}}{{\tan }^2\mathrm{x}+\frac{25}{9}}$$$$\mathrm{Put},\tan \mathrm{x}=\mathrm{t}$$$${\sec }^2\mathrm{xdx}=\mathrm{dt}$$$$\frac{1}{9}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+{\left(\frac{5}{3}\right)}^2}=\frac{1}{9}\times \frac{1}{\frac{5}{3}}{\tan }^{-1}\frac{\mathrm{t}}{\frac{5}{3}}+\mathrm{C}$$$$=\frac{3}{9\times 5}{\tan }^{-1}\frac{3\mathrm{t}}{5}+\mathrm{C}$$$$=\frac{1}{15}{\tan }^{-1}\frac{3\tan \mathrm{x}}{5}+\mathrm{C}//.$$

Answered by Dwaipayan Shikari last updated on 01/Aug/20

$$\int \frac{dx}{9+16{cos}^{2}x}=\int \frac{{sec}^{2}x}{9{sec}^{2}x+16}=\int \frac{{sec}^{2}x}{9{tan}^{2}x+25}dx$$$$\int \frac{dt}{9t^2+25}(tanx=t)$$$$\frac{1}{9}\int \frac{dt}{t^2+{\left(\frac{5}{3}\right)}^2}=\frac{1}{9}.\frac{3}{5}{tan}^{-1}\frac{3t}{5}+C=\frac{1}{15}{tan}^{-1}\frac{3tanx}{5}+C$$

Answered by mathmax by abdo last updated on 01/Aug/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{9+{16\cos }^2\mathrm{x}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{9+16\times \frac{1+\cos \left(2\mathrm{x}\right)}{2}}$$$$=\int \frac{\mathrm{dx}}{9+8+8\cos \left(2\mathrm{x}\right)}=\int \frac{\mathrm{dx}}{17+8\cos \left(2\mathrm{x}\right)}{=}_{2\mathrm{x}=\mathrm{t}}\int \frac{\mathrm{dt}}{2(17+8\mathrm{cost})}$$$$2\mathrm{I}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}}\int \frac{2\mathrm{du}}{(1+{\mathrm{u}}^2)(17+8\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2})}$$$$=\int \frac{2\mathrm{du}}{17+{17\mathrm{u}}^2+8-{8\mathrm{u}}^2}=\int \frac{2\mathrm{du}}{25+{9\mathrm{u}}^2}=\frac{2}{25}\int \frac{\mathrm{du}}{1+\frac{9}{25}{\mathrm{u}}^2}$$$${=}_{\mathrm{z}=\frac{3}{5}\mathrm{u}}\frac{2}{25}\int \frac{1}{1+{\mathrm{z}}^2}\times \frac{5}{3}\mathrm{dz}=\frac{2}{15}\mathrm{arctanz}+\mathrm{C}$$$$=\frac{2}{15}\arctan \left(\frac{3\mathrm{u}}{5}\right)+\mathrm{C}=\frac{2}{25}\arctan \left(\frac{3}{5}\tan \left(\frac{\mathrm{t}}{2}\right)\right)+\mathrm{C}$$$$=\frac{2}{15}\arctan \left(\frac{3}{5}\tan \left(\mathrm{x}\right)\right)+\mathrm{C}\Rightarrow$$$$\mathrm{I}=\frac{1}{15}\arctan \left(\frac{3}{5}\mathrm{tanx}\right)+\mathrm{C}$$

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