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Question Number 108921 by Ar Brandon last updated on 20/Aug/20

$$\mathrm{a}.\int \frac{{\sin }^{3}4\mathrm{x}}{{\cos }^{8}4\mathrm{x}}\mathrm{dx}$$$$\mathrm{b}.{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}({\mathrm{x}}^2{\mathrm{e}}^{\mathrm{cosx}}-2\mathrm{x})\mathrm{sinxdx}$$

Answered by bobhans last updated on 20/Aug/20

$$\frac{\approxeq \beta o\beta \cong }{\nparallel \nparallel \nparallel \nparallel }$$$$\left(a\right)\int \frac{(1-\cos {}^{2}4x).\sin 4x}{\cos {}^{8}4x}dx=-\frac{1}{4}\int \frac{(1-h^2)dh}{h^8}$$$$=-\frac{1}{4}\{\int h^{-8}dh-\int h^{-6}dh\}$$$$=-\frac{1}{4}\{-\frac{1}{7h^7}+\frac{1}{5h^5}\}+c$$$$=\frac{1}{28\cos {}^{7}4x}-\frac{1}{20\cos {}^{5}4x}+c$$

Commented by Ar Brandon last updated on 20/Aug/20

thanks

Answered by Ar Brandon last updated on 20/Aug/20

$$\mathrm{b}.\mathrm{I}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}({\mathrm{x}}^2{\mathrm{e}}^{\mathrm{cosx}}-2\mathrm{x})\mathrm{sinxdx},\mathrm{u}=-\mathrm{x}$$$$\mathrm{I}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}({\mathrm{x}}^2{\mathrm{e}}^{\mathrm{cosx}}+2\mathrm{x})\sin (-\mathrm{x})\mathrm{dx}$$$$2\mathrm{I}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}({\mathrm{x}}^2{\mathrm{e}}^{\mathrm{cosx}}-2\mathrm{x}-{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{cosx}}-2\mathrm{x})\mathrm{sinxdx}$$$$=-{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}4\mathrm{xsinxdx}$$$$\mathrm{I}=-{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}2\mathrm{xsinxdx}$$

Answered by Dwaipayan Shikari last updated on 20/Aug/20

$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(x^2{e}^{cosx}-2x)sinxdx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-(x^2{e}^{cosx}-2x)sinxdx=I$$$$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-4xsinxdx$$$$-\frac{I}{4}={\int }_0^{\frac{\pi }{2}}xsinxdx$$$$-\frac{I}{4}=-{\left[xcosx\right]}_0^{\frac{\pi }{2}}+{\left[sinx\right]}_0^{\frac{\pi }{2}}$$$$I=-4$$$$$$$$$$

Commented by Ar Brandon last updated on 20/Aug/20

Cool !

Commented by Ar Brandon last updated on 20/Aug/20

Hey bro, what do you think of Q108491 ? ��

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