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Question Number 103860 by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{find}\int \frac{\mathrm{dx}}{{\cos }^4\mathrm{x}}$$

Answered by Dwaipayan Shikari last updated on 17/Jul/20

$$\int \frac{dx}{{cos}^{4}x}=\int {sec}^{4}xdx=\int {sec}^{2}x(1+{tan}^{2}x)dx=\int (1+t^2)dt=t+\frac{t^3}{3}+C$$$$tanx+\frac{{tan}^{3}x}{3}+C\{putt=tanx$$

Answered by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{\cos }^4\mathrm{x}}\mathrm{we}\mathrm{know}1+{\tan }^2\mathrm{x}=\frac{1}{{\cos }^2\mathrm{x}}\Rightarrow \frac{1}{{\cos }^4\mathrm{x}}={(1+{\tan }^2\mathrm{x})}^2\Rightarrow$$$$\mathrm{I}=\int {(1+{\tan }^2\mathrm{x})}^2\mathrm{dx}{=}_{\mathrm{tanx}=\mathrm{u}}\int {(1+{\mathrm{u}}^2)}^2\frac{\mathrm{du}}{1+{\mathrm{u}}^2}=\int (1+{\mathrm{u}}^2)\mathrm{du}$$$$=\mathrm{u}+\frac{{\mathrm{u}}^3}{3}+\mathrm{C}=\mathrm{tanx}+\frac{1}{3}{\tan }^3\mathrm{x}+\mathrm{C}.$$

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