यदि int operatorname{cosec}^5 x , dx = alpha | vedclass

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(B) माना $I = \int \operatorname{cosec}^5 x \, dx$. खंडशः समाकलन (integration by parts) का उपयोग करते हुए,$u = \operatorname{cosec}^3 x$ और $dv = \ope

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$1$

Vedclass · Answer

(B) माना $I = \int \operatorname{cosec}^5 x \, dx$. खंडशः समाकलन (integration by parts) का उपयोग करते हुए,$u = \operatorname{cosec}^3 x$ और $dv = \operatorname{cosec}^2 x \, dx$ लेने पर:$I = \operatorname{cosec}^3 x(-\cot x) - \int (3 \operatorname{cosec}^2 x(-\operatorname{cosec} x \cot x))(-\cot x) \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x \cot^2 x \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x (\operatorname{cosec}^2 x - 1) \, dx$$I = -\operatorname{cosec}^3 x \cot x - 3I + 3 \int \operatorname{cosec}^3 x \, dx$$4I = -\operatorname{cosec}^3 x \cot x + 3 \int \operatorname{cosec}^3 x \, dx$हम जानते हैं कि $\int \operatorname{cosec}^3 x \, dx = -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \ln |	an \frac{x}{2}|$.इस मान को $4I$ के समीकरण में रखने पर:$4I = -\operatorname{cosec}^3 x \cot x + 3 \left( -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \ln |	an \frac{x}{2}| ight)$$4I = -\operatorname{cosec}^3 x \cot x - \frac{3}{2} \operatorname{cosec} x \cot x + \frac{3}{2} \ln |	an \frac{x}{2}|$$I = -\frac{1}{4} \cot x \operatorname{cosec} x (\operatorname{cosec}^2 x + \frac{3}{2}) + \frac{3}{8} \ln |	an \frac{x}{2}| + C$दिए गए रूप से तुलना करने पर,$\alpha = -\frac{1}{4}$ और $\beta = \frac{3}{8}$.अतः,$8(\alpha + \beta) = 8(-\frac{1}{4} + \frac{3}{8}) = 8(\frac{-2+3}{8}) = 1$.

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