int {frac{{cos 2x + 2{{sin }^2}x}}{{{{cos }^2 | vedclass

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Question

(C) આપણે જાણીએ છીએ કે $\cos 2x = 1 - 2\sin^2 x$. આ કિંમત સંકલનમાં મૂકતા:$\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} = \int {\frac{{(1 -

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$	an x + c$

Vedclass · Answer

(C) આપણે જાણીએ છીએ કે $\cos 2x = 1 - 2\sin^2 x$. આ કિંમત સંકલનમાં મૂકતા:$\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} = \int {\frac{{(1 - 2{{\sin }^2}x) + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx}$$= \int {\frac{1}{{{{\cos }^2}x}}dx}$$= \int {{{\sec }^2}x\,dx}$$= 	an x + c$

$\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} = $

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