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(D) ધારો કે $I = \int \frac{1}{\cos x} \left[ \frac{\sin x + 3 \cos x - \sin x}{\sin x(\sin x + 3 \cos x)} \right] dx$ $= \int \frac{1}{\cos x} \left[

Vedclass · Accepted Answer

$\log \left| \frac{\sin x}{\sin x + 3 \cos x} \right| + c$

Vedclass · Answer

(D) ધારો કે $I = \int \frac{1}{\cos x} \left[ \frac{\sin x + 3 \cos x - \sin x}{\sin x(\sin x + 3 \cos x)} ight] dx$ $= \int \frac{1}{\cos x} \left[ \frac{3 \cos x}{\sin x(\sin x + 3 \cos x)} ight] dx$ $= \int \frac{3}{\sin x(\sin x + 3 \cos x)} dx$ અંશ અને છેદને $\cos^2 x$ વડે ભાગતા: $= \int \frac{3 \sec^2 x}{	an x(	an x + 3)} dx$ ધારો કે $u = 	an x$,તેથી $du = \sec^2 x dx$. $I = \int \frac{3}{u(u + 3)} du$ આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા: $\frac{3}{u(u + 3)} = \frac{A}{u} + \frac{B}{u + 3} \implies 3 = A(u + 3) + Bu$. $u = 0$ માટે,$A = 1$. $u = -3$ માટે,$B = -1$. $I = \int \left( \frac{1}{u} - \frac{1}{u + 3} ight) du = \log |u| - \log |u + 3| + c$ $= \log \left| \frac{u}{u + 3} ight| + c = \log \left| \frac{	an x}{	an x + 3} ight| + c$ $= \log \left| \frac{\sin x / \cos x}{(\sin x + 3 \cos x) / \cos x} ight| + c = \log \left| \frac{\sin x}{\sin x + 3 \cos x} ight| + c$.

$\int \frac{1}{\cos x} \left[ \frac{1}{\sin x} - \frac{1}{\sin x + 3 \cos x} \right] dx =$

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