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Question Number 45795 by maxmathsup by imad last updated on 16/Oct/18

$$find\int \frac{dx}{cosx{sin}^{2}x}$$

Answered by MJS last updated on 17/Oct/18

$$\frac{1}{\cos x{\sin }^{2}x}=(1+\frac{{\cos }^{2}x}{{\sin }^{2}x})\frac{1}{\cos x}=$$$$=\frac{1}{\cos x}+\frac{\cos x}{{\sin }^{2}x}$$$$\int \frac{dx}{\cos x}=\int \sec xdx=\ln \mid \tan x+\sec x\mid =\ln \mid \frac{1+\sin x}{\cos x}\mid$$$$\int \frac{\cos x}{{\sin }^{2}x}dx=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=\int \frac{dt}{t^2}=-\frac{1}{t}=-\frac{1}{\sin x}$$$$\int \frac{dx}{\cos x{\sin }^{2}x}=\ln \mid \frac{1+\sin x}{\cos x}\mid -\frac{1}{\sin x}+C$$

Commented by math khazana by abdo last updated on 17/Oct/18

$$thankyousirMJS.$$

Commented by MJS last updated on 17/Oct/18

$$\mathrm{btw}$$$$\frac{1}{{\cos }^{2}x\sin x}=(1+\frac{{\sin }^{2}x}{{\cos }^{2}x})\frac{1}{\sin x}=$$$$=\frac{1}{\sin x}+\frac{\sin x}{{\cos }^{2}x}$$$$\int \frac{dx}{\sin x}=\int \csc xdx=-\ln \mid \cot x+\csc x\mid =-\ln \mid \frac{1+\cos x}{\sin x}\mid$$$$\int \frac{\sin x}{{\cos }^{2}x}dx=$$$$[t=\cos x\to dx=-\frac{dt}{\sin x}]$$$$=-\int \frac{dt}{t^2}=\frac{1}{t}=\frac{1}{\cos x}$$

Commented by MJS last updated on 17/Oct/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}.\mathrm{I}\mathrm{just}\mathrm{fixed}\mathrm{a}\mathrm{typo}...$$

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