निम्नलिखित समाकलन int_{frac{pi}{4}}^{frac{pi} | vedclass

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(A) माना $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \operatorname{cosec} x)^{17} dx$.$	an(\frac{x}{2}) = e^u$ प्रतिस्थापित करने पर,$\frac{1}{2} \se

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$\int_0^{\log (1+\sqrt{2})} 2(e^u+e^{-u})^{16} du$

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(A) माना $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \operatorname{cosec} x)^{17} dx$.$	an(\frac{x}{2}) = e^u$ प्रतिस्थापित करने पर,$\frac{1}{2} \sec^2(\frac{x}{2}) dx = e^u du$,अतः $dx = \frac{2 e^u}{1+e^{2u}} du$.साथ ही,$\operatorname{cosec} x = \frac{1}{\sin x} = \frac{1+e^{2u}}{2e^u} = \frac{e^u + e^{-u}}{2}$.जब $x = \frac{\pi}{4}$,तो $e^u = 	an(\frac{\pi}{8}) = \sqrt{2}-1$,अतः $u = \ln(\sqrt{2}-1) = -\ln(\sqrt{2}+1)$.जब $x = \frac{\pi}{2}$,तो $e^u = 	an(\frac{\pi}{4}) = 1$,अतः $u = 0$.इन मानों को समाकलन में रखने पर:$I = \int_{-\ln(\sqrt{2}+1)}^{0} (e^u + e^{-u})^{17} \cdot \frac{2}{e^u + e^{-u}} du = \int_{-\ln(\sqrt{2}+1)}^{0} 2(e^u + e^{-u})^{16} du$.चूंकि $f(u) = (e^u + e^{-u})^{16}$ एक सम फलन है,इसलिए $\int_{-a}^{0} f(u) du = \int_{0}^{a} f(u) du$.अतः,$I = \int_{0}^{\ln(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$.

निम्नलिखित समाकलन $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \operatorname{cosec} x)^{17} d x$ किसके बराबर है?

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