Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 93958 by mashallah last updated on 16/May/20

$$\int \frac{\mathrm{sinx}-\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=$$

Commented by mathmax by abdo last updated on 17/May/20

$$I=\int \frac{sinx-cosx}{sinx+cosx}dx\Rightarrow I=\int \frac{tanx-1}{tanx+1}dx{=}_{tanx=t}$$$$=\int \frac{t-1}{t+1}\times \frac{dt}{1+t^2}=\int \frac{t-1}{(t+1)(t^2+1)}dtletdecompose$$$$F\left(t\right)=\frac{t-1}{(t+1)(t^2+1)}\Rightarrow F\left(t\right)=\frac{a}{t+1}+\frac{bt+c}{t^2+1}$$$$a=-1{lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=a+b\Rightarrow b=1$$$$F\left(o\right)=-1=a+c\Rightarrow c=0\Rightarrow F\left(t\right)=-\frac{1}{t+1}+\frac{t}{t^2+1}\Rightarrow$$$$I=-ln\mid t+1\mid +\frac{1}{2}ln(1+t^2)+C$$$$=\frac{1}{2}ln(1+{tan}^{2}x)-ln\mid 1+tanx\mid +C$$

Answered by john santu last updated on 16/May/20

$$\frac{{(\sin \mathrm{x}-\cos \mathrm{x})}^2}{-\cos 2\mathrm{x}}=\frac{1-\sin 2\mathrm{x}}{-\cos 2\mathrm{x}}$$$$=\tan 2\mathrm{x}-\sec 2\mathrm{x}$$$$$$

Answered by Kunal12588 last updated on 16/May/20

$$sinx+cosx=t$$$$\Rightarrow -(sinx-cosx)dx=dt$$$$I=\int -\frac{dt}{t}=-ln\mid sinx+cosx\mid$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com