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Question Number 187408 by MathsFan last updated on 17/Feb/23

$$\int \frac{{\mathbf{sec}}^2\mathbf{x}}{\sqrt{1-{\mathbf{tan}}^2\mathbf{x}}}\mathbf{dx}$$

Answered by horsebrand11 last updated on 17/Feb/23

$$=\int \frac{d\left(\tan x\right)}{\sqrt{1-\tan {}^{2}x}};u=\tan x$$$$=\int \frac{du}{\sqrt{1-u^2}};u=\sin t$$$$=\int \frac{\cos t}{\sqrt{1-\sin {}^{2}t}}dt=t+C$$$$=\arcsin \left(\tan x\right)+C$$

Commented by MJS_new last updated on 17/Feb/23

$$\arcsin \tan x$$$$\mathrm{check}\mathrm{it}!$$

Commented by horsebrand11 last updated on 17/Feb/23

$$yesthanks$$

Answered by MJS_new last updated on 17/Feb/23

$$\mathrm{I}\mathrm{love}\mathrm{complicated}\mathrm{paths}$$$$\int \frac{{\sec }^{2}x}{\sqrt{1-{\tan }^{2}x}}dx=\int \frac{dx}{\cos x\sqrt{1-{2\sin }^{2}x}}=$$$$[t=\frac{\sqrt{2}\sin x}{\sqrt{1-{2\sin }^{2}x}}\to dx=\frac{{(1-{2\sin }^{2}x)}^{3/2}}{\sqrt{2}\cos x}]$$$$=\sqrt{2}\int \frac{dt}{t^2+2}=\arctan \frac{t}{\sqrt{2}}=$$$$=\arctan \frac{\sin x}{\sqrt{1-{2\sin }^{2}x}}=\arcsin \tan x+C$$

Commented by MathsFan last updated on 17/Feb/23

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$$$\mathrm{i}\mathrm{appreciate}\mathrm{your}\mathrm{effort}$$

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