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Question Number 172420 by Physicien last updated on 26/Jun/22

Answered by thfchristopher last updated on 26/Jun/22

$$\int \frac{1}{7+\tan x}dx$$$$\mathrm{Let}t=\tan \frac{x}{2}$$$$dt=\frac{1}{2}{\sec }^{2}xdx$$$$=\frac{1}{2}(1+t^2)dc$$$$\Rightarrow dx=\frac{2dt}{1+t^2}$$$$\tan x=\frac{2t}{1-t^2},\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$$$$\therefore \int \frac{1}{7+\tan x}dx$$$$=\int \frac{\frac{2dt}{1+t^2}}{7+\frac{2t}{1-t^2}}$$$$=\int \frac{2(1-t^2)dt}{(1+t^2)(7-7t^2+2t)}$$$$\frac{2t^2-2}{(t^2+1)(7t^2-2t-7)}=\frac{At+B}{t^2+1}+\frac{Ct+D}{7t^2-2t-7}$$$$\mathrm{By}\mathrm{solving},$$$$A=-\frac{1}{25},B=\frac{7}{25},C=\frac{7}{25},D=-\frac{1}{25}$$$$\frac{2t^2-2}{(t^2+1)(7t^2-2t-7)}=\frac{1}{25}(\frac{7-t}{t^2+1}+\frac{7t-1}{7t^2-2t-7})$$$$\int \frac{2(1-t^2)dt}{(1+t^2)(7-7t^2+2t)}$$$$=\frac{1}{25}(\int \frac{7-t}{t^2+1}dt+\int \frac{7t-1}{7t^2-2t-7}dt)$$$$=\frac{1}{50}[-\int \frac{2t-14}{t^2+1}dt-\int \frac{14t-2}{7t^2-2t-7}dt]$$$$=\frac{1}{50}[14\int \frac{dt}{t^2+1}-\ln \mid t^2+1\mid +\ln \mid 7t^2-2t-7\mid ]$$$$=\frac{7}{25}{\tan }^{-1}t+\frac{1}{50}\ln \mid \frac{7t^2-2t-7}{t^2+1}\mid +C$$$$=\frac{7x}{50}+\frac{1}{50}\ln \mid -\sin x-7\cos x\mid +C$$$$=\frac{1}{50}(7x+\ln \mid \sin x+7\cos x\mid )+C$$

Commented by Tawa11 last updated on 26/Jun/22

$$\mathrm{Great}\mathrm{sir}.$$

Commented by peter frank last updated on 03/Jul/22

$$\mathrm{thank}\mathrm{you}$$

Answered by MJS_new last updated on 26/Jun/22

$$\int \frac{dx}{7+\tan x}=$$$$[t=\tan x\to dx={\cos }^{2}xdt=\frac{dt}{t^2+1}]$$$$=\int \frac{dt}{(t+7)(t^2+1)}=$$$$=\frac{1}{50}\int \frac{dt}{t+7}-\frac{1}{100}\int \frac{2t}{t^2+1}dt+\frac{7}{50}\int \frac{dt}{t^2+1}=$$$$=\frac{1}{50}\ln (t+7)-\frac{1}{100}\ln (t^2+1)+\frac{7}{50}\arctan t=$$$$=\frac{1}{100}\ln \frac{{(t+7)}^2}{t^2+1}+\frac{7}{50}\arctan t=$$$$=\frac{7}{50}x+\frac{1}{50}\ln \mid 7\cos x+\sin x\mid +C$$

Commented by thfchristopher last updated on 26/Jun/22

$$\mathrm{Your}\mathrm{solution}\mathrm{is}\mathrm{simple}\mathrm{than}\mathrm{mine}.\mathrm{Thanks}!$$

Commented by MJS_new last updated on 26/Jun/22

$$\mathrm{in}\mathrm{some}\mathrm{cases}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{need}t=\tan \frac{x}{2}\mathrm{because}$$$$t=\tan x\mathrm{works}\mathrm{too}.\mathrm{To}\mathrm{see}\mathrm{this}\mathrm{needs}\mathrm{just}\mathrm{a}$$$$\mathrm{little}\mathrm{experience}.$$

Commented by Tawa11 last updated on 26/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

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