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Question Number 146597 by iloveisrael last updated on 14/Jul/21

$$\int \frac{\cos 5\mathrm{x}+\cos 4\mathrm{x}}{1-2\cos 3\mathrm{x}}\mathrm{dx}=?$$$$\int \frac{\sqrt{\tan \mathrm{x}}}{\sin 2\mathrm{x}}\mathrm{dx}=?$$$$\int \frac{\mathrm{dx}}{\sqrt{\cos {}^3\mathrm{x}\sin {}^5\mathrm{x}}}=?$$

Answered by gsk2684 last updated on 14/Jul/21

$$\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$$$$\int \frac{2\cos \frac{5x+4x}{2}\cos \frac{5x-4x}{2}}{1-2(2\cos {}^2\frac{3x}{2}-1)}dx$$$$2\int \frac{\cos 3\left(\frac{3x}{2}\right)\cos \frac{x}{2}}{1-4\cos {}^2\frac{3x}{2}+2}dx$$$$2\int \frac{[4\cos {}^3\left(\frac{3x}{2}\right)-3\cos \left(\frac{3x}{2}\right)]\cos \frac{x}{2}}{3-4\cos {}^2\frac{3x}{2}}dx$$$$2\int \frac{[4\cos {}^2\left(\frac{3x}{2}\right)-3]\cos \frac{3x}{2}\cos \frac{x}{2}}{3-4\cos {}^2\frac{3x}{2}}dx$$$$-\int 2\cos \frac{3x}{2}\cos \frac{x}{2}dx$$$$-\int (\cos (\frac{3x}{2}+\frac{x}{2})+\cos (\frac{3x}{2}-\frac{x}{2}))dx$$$$-\int (\cos 2x+\cos x)dx$$$$-(\frac{\sin 2x}{2}+\sin x)+c$$

Answered by gsk2684 last updated on 14/Jul/21

$$\int \frac{dx}{\sqrt{\cos {}^{3}x\sin {}^{5}x}}dx$$$$\int \frac{1}{\cos {}^{4}x\sqrt{\frac{\cos {}^{3}x\sin {}^{5}x}{\cos {}^{8}x}}}dx$$$$\int \frac{\sec {}^{2}x}{\sqrt{\tan {}^{5}x}}\sec {}^{2}xdx$$$$\int \frac{1+\tan {}^{2}x}{\sqrt{\tan {}^{5}x}}d\left(\tan x\right)$$$$\int (\tan {}^{-\frac{5}{2}}x+\frac{1}{\sqrt{\tan x}})d\left(\tan x\right)$$$$\frac{\tan {}^{-\frac{5}{2}+1}x}{-\frac{5}{2}+1}+2\sqrt{\tan x}+c$$

Answered by gsk2684 last updated on 14/Jul/21

$$\int \frac{\sqrt{\tan x}}{\sin 2x}dx$$$$\int \frac{\sqrt{\tan x}}{\left(\frac{2\tan x}{1+\tan {}^{2}x}\right)}dx$$$$=\int \frac{1}{2\sqrt{\tan x}}\sec {}^{2}xdx$$$$=\int \frac{1}{2\sqrt{\tan x}}d\left(\tan x\right)$$$$=\sqrt{\tan x}+c$$

Answered by puissant last updated on 14/Jul/21

$$2)$$$$\int \frac{\sqrt{\tan \mathrm{x}}}{\sin 2\mathrm{x}}\mathrm{dx}=\mathrm{K}$$$$\mathrm{u}=\sqrt{\tan \mathrm{x}}\Rightarrow {\mathrm{u}}^2=\tan \mathrm{x}2\mathrm{udu}=\frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}$$$$\Rightarrow \mathrm{dx}={2\mathrm{ucos}}^2\mathrm{x}\mathrm{dx}$$$$\mathrm{K}=\int \frac{\mathrm{u}}{2\sin \mathrm{x}\cos \mathrm{x}}\times {2\mathrm{ucos}}^2\mathrm{x}\mathrm{du}$$$$=\int \frac{{\mathrm{u}}^2}{\tan \mathrm{x}}\mathrm{du}=\int \frac{{\mathrm{u}}^2}{{\mathrm{u}}^2}\mathrm{du}=\mathrm{u}+\mathrm{c}$$$$\Rightarrow \mathrm{K}=\sqrt{\tan \mathrm{x}}+\mathrm{c}..$$

Answered by mathmax by abdo last updated on 14/Jul/21

$$\mathrm{I}=\int \frac{\sqrt{\mathrm{tanx}}}{\sin \left(2\mathrm{x}\right)}\mathrm{dx}\mathrm{changement}\mathrm{tanx}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}=\int \frac{\sqrt{\mathrm{t}}}{\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}=\frac{1}{2}\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\sqrt{\mathrm{t}}+\mathrm{C}=\sqrt{\mathrm{tanx}}+\mathrm{C}$$

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