यदि int frac{sin x}{sin 4x} dx = alpha log le | vedclass

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(D) दिया गया है $I = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx = \frac{1}{4} \int \frac{dx}{\

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(D) दिया गया है $I = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx = \frac{1}{4} \int \frac{dx}{\cos x (1 - 2 \sin^2 x)}$।अंश और हर को $\cos x$ से गुणा करने पर: $I = \frac{1}{4} \int \frac{\cos x dx}{\cos^2 x (1 - 2 \sin^2 x)} = \frac{1}{4} \int \frac{\cos x dx}{(1 - \sin^2 x)(1 - 2 \sin^2 x)}$।माना $u = \sin x$, तब $du = \cos x dx$। $I = \frac{1}{4} \int \frac{du}{(1 - u^2)(1 - 2u^2)}$।आंशिक भिन्नों का उपयोग करने पर: $\frac{1}{(1 - u^2)(1 - 2u^2)} = \frac{A}{1 - u^2} + \frac{B}{1 - 2u^2}$।$1 = A(1 - 2u^2) + B(1 - u^2)$। $u^2 = 1$ के लिए, $A = -1$। $u^2 = 1/2$ के लिए, $B = 2$।$I = \frac{1}{4} \int \left( \frac{-1}{1 - u^2} + \frac{2}{1 - 2u^2} \right) du = -\frac{1}{4} \int \frac{du}{1 - u^2} + \frac{2}{4} \int \frac{du}{1 - (\sqrt{2}u)^2}$।$I = -\frac{1}{8} \log \left| \frac{1+u}{1-u} \right| + \frac{1}{2} \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2}u}{1-\sqrt{2}u} \right| + c$।दिए गए रूप से तुलना करने पर: $\alpha = -1/8$, $\beta = 1/(4\sqrt{2})$।$\beta^2 = 1/32$। अतः, $32(\alpha + \beta^2) = 32(-1/8 + 1/32) = 32(-4/32 + 1/32) = 32(-3/32) = -3$।

यदि $\int \frac{\sin x}{\sin 4x} dx = \alpha \log \left| \frac{1+\sin x}{1-\sin x} \right| + \beta \log \left| \frac{1+\sqrt{2} \sin x}{1-\sqrt{2} \sin x} \right| + c$ है, तो $32(\alpha + \beta^2) = $ का मान क्या है?

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