int tan ^{-1}(sec x+tan x) d x= | vedclass

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(A) Let $I = \int 	an ^{-1}(\sec x+	an x) d x$.We know that $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\cos x}$,so $\sec x + 	an x = \

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$\frac{\pi x}{4}+\frac{x^2}{4}+c$

Vedclass · Answer

(A) Let $I = \int 	an ^{-1}(\sec x+	an x) d x$.We know that $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\cos x}$,so $\sec x + 	an x = \frac{1+\sin x}{\cos x}$.Using half-angle formulas,$1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = (\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})$.Thus,$\frac{1+\sin x}{\cos x} = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}$.Dividing numerator and denominator by $\cos \frac{x}{2}$,we get $\frac{1+	an \frac{x}{2}}{1-	an \frac{x}{2}} = 	an(\frac{\pi}{4} + \frac{x}{2})$.Therefore,$I = \int 	an^{-1}(	an(\frac{\pi}{4} + \frac{x}{2})) d x = \int (\frac{\pi}{4} + \frac{x}{2}) d x$.Integrating term by term,we get $I = \frac{\pi x}{4} + \frac{x^2}{4} + c$.

$\int \tan ^{-1}(\sec x+\tan x) d x=$

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