Question

Evaluate: $\int \sqrt{1+\sec x}dx$

• A. $-{\cos }^{-1}(2\cos x+1)+c$
• B. $-{\tan }^{-1}(2\cos x+1)+c$
• C. $-{\cot }^{-1}(2\cos x-1)+c$
• D. $-{\sin }^{-1}(2\cos x-1)+c$

Answer 1

Answer: (D)

$-{\sin }^{-1}(2\cos x-1)+c$

$I=\int \sqrt{1+\sec x}dx$

$=\int \sqrt{1+\frac{1}{\cos x}}dx$

$=\int \sqrt{\frac{1+\cos x}{\cos x}}dx$

$=\int \sqrt{\frac{1-{\cos }^{2}x}{\cos x(1-\cos }}dx$

$=\int \frac{\sin x}{\sqrt{\cos x(1-\cos x)}}dx$

Take $\cos x=t$ $\Rightarrow -\sin xdx=dt$

$\Rightarrow I=-\int dt\sqrt{t(1-t)}$

$=-\int \frac{dt}{\sqrt{t-t^2}}$

$=-\int \frac{dt}{\sqrt{t-t^2}+\frac{1}{4}-\frac{1}{4}}$

$=-\int \frac{dt}{\sqrt{(\frac{1}{2}{)}^2-(t-\frac{1}{2}{)}^2}}$

$=-{\sin }^{-1}\frac{t-\frac{1}{2}}{\frac{1}{2}}$

$=-{\sin }^{-1}(2t-1)+c$

$=-{\sin }^{-1}(2\cos x-1)+c$