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Question Number 128349 by mathocean1 last updated on 06/Jan/21

$$functiongisdefinedin[0;\frac{\pi }{4}]$$$$byg\left(x\right)=\frac{sinx}{{cos}^{3}x}.$$$$1.Determinatea;b\in \mathbb{R}suchthat$$$$g^{\prime }\left(x\right)=\frac{a}{{cos}^{4}x}+\frac{b}{{cos}^{2}x}.$$$$2.DeducttheprimitiveGofthe$$$$functiont\left(x\right)=\frac{1}{{cos}^{4}x}suchthat$$$$t\left(\frac{\pi }{4}\right)=1.$$

Answered by mathmax by abdo last updated on 06/Jan/21

$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}}{{\cos }^3\mathrm{x}}\Rightarrow {\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{{\cos }^4\mathrm{x}-{3\mathrm{cosx}}^2(-\mathrm{sinx})\mathrm{sinx}}{{\cos }^6\mathrm{x}}$$$$=\frac{{\cos }^4\mathrm{x}+{3\cos }^2\mathrm{x}{\sin }^2\mathrm{x}}{{\cos }^6\mathrm{x}}=\frac{1}{{\cos }^2\mathrm{x}}+\frac{{3\sin }^2\mathrm{x}}{{\cos }^4\mathrm{x}}$$$$=\frac{1}{{\cos }^2\mathrm{x}}+\frac{3(1-{\cos }^2\mathrm{x})}{{\cos }^4\mathrm{x}}=\frac{1}{{\cos }^2\mathrm{x}}-\frac{3}{{\cos }^2\mathrm{x}}+\frac{3}{{\cos }^4\mathrm{x}}=\frac{-2}{{\cos }^2\mathrm{x}}+\frac{3}{{\cos }^4\mathrm{x}}\Rightarrow$$$$\int {\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)\mathrm{dx}=-2\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}+3\int \frac{\mathrm{dx}}{{\cos }^4\mathrm{x}}\Rightarrow$$$$\mathrm{g}\left(\mathrm{x}\right)=-2\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}+3\int \frac{\mathrm{dx}}{{\cos }^4\mathrm{x}}\Rightarrow$$$$3\int \frac{\mathrm{dx}}{{\cos }^4\mathrm{x}}=\frac{\mathrm{sinx}}{{\cos }^3\mathrm{x}}+2\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}\mathrm{we}\mathrm{have}$$$$\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}=\int \frac{2\mathrm{dx}}{1+\cos \left(2\mathrm{x}\right)}{=}_{\mathrm{tanx}=\mathrm{t}}2\int \frac{\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$=2\int \frac{\mathrm{dt}}{1+{\mathrm{t}}^2+1-{\mathrm{t}}^2}=\mathrm{t}+\mathrm{c}=\mathrm{tanx}+\mathrm{c}\Rightarrow$$$$\int \frac{\mathrm{dx}}{{\cos }^4\mathrm{x}}=\frac{\mathrm{sinx}}{{3\cos }^3\mathrm{x}}+\frac{2}{3}\mathrm{tanx}+\mathrm{C}$$

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