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

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(B) ધારો કે $I = \int \operatorname{cosec}^5 x \, dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા,$u = \operatorname{cosec}^3 x$ અને $dv = \operatorname{cosec}^2 x \, d

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(B) ધારો કે $I = \int \operatorname{cosec}^5 x \, dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા,$u = \operatorname{cosec}^3 x$ અને $dv = \operatorname{cosec}^2 x \, dx$ લેતા:$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$આપણે જાણીએ છીએ કે $\int \operatorname{cosec}^3 x \, dx = -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \ln |	an \frac{x}{2}|$.આ કિંમત $4I$ ના સમીકરણમાં મૂકતા:$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$આપેલ સ્વરૂપ સાથે સરખાવતા,$\alpha = -\frac{1}{4}$ અને $\beta = \frac{3}{8}$.તેથી,$8(\alpha + \beta) = 8(-\frac{1}{4} + \frac{3}{8}) = 8(\frac{-2+3}{8}) = 1$.

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