int tan ^{-1}left(sqrt{frac{1-sin x}{1+sin x} | vedclass

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Question

(A) ધારો કે $I = \int 	an ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}ight) d x$ નિત્યસમ $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ અને $1

Vedclass · Accepted Answer

$\frac{\pi}{4} x-\frac{x^2}{4}+c$,જ્યાં $c$ એ સંકલનનો અચળાંક છે.

Vedclass · Answer

(A) ધારો કે $I = \int 	an ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}ight) d x$ નિત્યસમ $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ અને $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ નો ઉપયોગ કરતા: $I = \int 	an ^{-1}\left(\sqrt{\frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}}ight) d x$ $I = \int 	an ^{-1}\left(\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}ight) d x$ અંશ અને છેદને $\cos \frac{x}{2}$ વડે ભાગતા: $I = \int 	an ^{-1}\left(\frac{1 - 	an \frac{x}{2}}{1 + 	an \frac{x}{2}}ight) d x$ કારણ કે $	an(\frac{\pi}{4} - 	heta) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$: $I = \int 	an ^{-1}\left(	an \left(\frac{\pi}{4} - \frac{x}{2}ight)ight) d x$ $I = \int \left(\frac{\pi}{4} - \frac{x}{2}ight) d x$ $I = \frac{\pi}{4} x - \frac{x^2}{4} + c$

$\int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x=$

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