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 36738 by MJS last updated on 04/Jun/18

$$\left(1\right)\int \frac{d\alpha }{{(1+\sin 2\alpha )}^2}=$$$$\left(2\right)\int \frac{d\beta }{{(1+\cos 2\beta )}^2}=$$$$\left(3\right)\int \frac{d\gamma }{(1+\sin 2\gamma )(1+\cos 2\gamma )}=$$

Commented by behi83417@gmail.com last updated on 05/Jun/18

$$sin2\alpha =\frac{e^{2\alpha }-e^{-2\alpha }}{2i}=\frac{e^{4\alpha }-1}{2{ie}^{2\alpha }}=\frac{x^4-1}{2{ix}^2}$$$$\Rightarrow I=\int \frac{dx}{{(1+\frac{x^4-1}{2{ix}^2})}^2}=\int \frac{-4x^{4}dx}{{(x^4+2{ix}^2-1)}^2}=$$$$=\int \frac{-4x^{4}dx}{{(x^2+i)}^4}=-4\int \frac{dx}{{(x+\frac{i}{x})}^4}=....$$

Commented by abdo.msup.com last updated on 05/Jun/18

$$1)I=\int \frac{dx}{{(1+sin\left(2x\right))}^2}$$$${=}_{2x=t}\int \frac{1}{{(1+sint)}^2}\frac{dt}{2}$$$$2I{=}_{tan\left(\frac{t}{2}\right)=u}\int \frac{1}{{(1+\frac{2u}{1+u^2})}^2}\frac{2du}{1+u^2}$$$$=2\int \frac{{(1+u^2)}^2}{{(1+u^2+2u)}^2}\frac{du}{1+u^2}$$$$=2\int \frac{1+u^2}{{(u+1)}^4}duletdecompose$$$${=}_{u+1=x}2\int \frac{1+{(x-1)}^2}{x^4}dx$$$$=2\int \frac{1+x^2-2x+1}{x^4}dx$$$$I=\int (\frac{2}{x^4}+\frac{1}{x^2}-\frac{2}{x^3})dx$$$$=2.\frac{1}{-4+1}{x}^{-4+1}-\frac{1}{x}-2\frac{1}{-3+1}{x}^{-3+1}+c$$$$=\frac{-2}{3x^3}-\frac{1}{x}+\frac{1}{x^2}+c$$$$=\frac{-2}{3{(u+1)}^3}-\frac{1}{u+1}+\frac{1}{{(u+1)}^2}+c$$$$=\frac{-2}{3{(1+tanx)}^3}-\frac{1}{1+tanx}+\frac{1}{{(1+tanx)}^2}+c$$

Commented by abdo.msup.com last updated on 05/Jun/18

$$3)letI=\int \frac{dx}{(1+sin\left(2x\right))(1+cos\left(2x\right))}$$$$I{=}_{2x=t}\int \frac{1}{(1+sint)(1+cost)}\frac{dt}{2}$$$$2I{=}_{tan\left(\frac{t}{2}\right)=u}\int \frac{1}{(1+\frac{2u}{1+u^2})(1+\frac{1-u^2}{1+u^2})}\frac{2du}{1+u^2}$$$$=\int \frac{2du}{(1+u^2)\left(\frac{1+u^2+2u}{1+u^2}\right)\left(\frac{1+u^2+1-u^2}{1+u^2}\right)}$$$$I=\int \frac{1+u^2}{2{(u+1)}^2}du$$$$2I{=}_{u+1=x}\int \frac{1+{(x-1)}^2}{x^2}dx$$$$=\int \frac{1+x^2-2x+1}{x^2}dx$$$$=\int (\frac{2}{x^2}+1-\frac{2}{x})dx$$$$=-\frac{2}{x}+x-2ln\mid x\mid +c$$$$=-\frac{2}{u+1}+u+1-2ln\mid u+1\mid +c$$$$=-\frac{2}{1+tan\left(x\right)}+1+tan\left(x\right)-2ln\mid 1+tanx\mid +c$$$$I=\frac{-1}{1+tanx}+\frac{1}{2}tanx-ln\mid 1+tanx\mid +\lambda$$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

$$3)t=tan\gamma dt={sec}^2\gamma d\gamma$$$$\frac{dt}{1+t^2}=d\gamma$$$$\int \frac{dt}{1+t^2}\times \frac{1}{(1+\frac{2t}{1+t^2})(1+\frac{1-t^2}{1+t^2})}$$$$\int \frac{1+t^{2}dt}{(1+t^2+2t)\left(2\right)}$$$$=\frac{1}{2}\int \frac{1+t^2}{1+t^2+2t}$$$$=\frac{1}{2}\int \frac{1+t^2+2t-2t}{1+t^2+2t}dt$$$$=\frac{1}{2}\int dt-\frac{1}{2}\int \frac{2t+2-2}{1+t^2+2t}$$$$=\frac{1}{2}\int dt-\frac{1}{2}\int \frac{d(1+t^2+2t)}{1+t^2+2t}+\int \frac{dt}{{(1+t)}^2}$$$$=\frac{1}{2}t-\frac{1}{2}ln(1+t^2+2t)+\frac{{(1+t)}^{-1}}{-1}+c$$$$=\frac{1}{2}tan\gamma -\frac{1}{2}\times 2ln(1+t)-\frac{1}{1+t}+c$$$$=\frac{1}{2}tan\gamma -ln(1+tan\gamma )-\frac{1}{1+tan\gamma }+c$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

$$1)tan\alpha =tdt={sec}^2\alpha d\alpha$$$$d\alpha =\frac{dt}{1+t^2}$$$$\int \frac{dt}{1+t^2}\times \frac{{(1+t^2)}^2}{{(1+t^2+2t)}^2}$$$$\int \frac{(1+t^2)dt}{{(1+t^2+2t)}^2}$$$$\int \frac{1+t^2+2t-2t-2+2}{{(1+t^2+2t)}^2}dt$$$$\int \frac{dt}{1+t^2+2t}-\int \frac{d(t^2+2t+1)}{{(1+t^2+2t)}^2}+2\int \frac{dt}{{(1+t)}^4}$$$$\int \frac{dt}{{(1+t)}^2}-\int \frac{d(t^2+2t+1)}{{(1+t^2+2t)}^2}+2\int \frac{dt}{{(1+t)}^4}$$$$=\frac{{(1+t)}^{-1}}{-1}-\frac{{(1+2t+t^2)}^{-1}}{-1}+2\frac{{(1+t)}^{-3}}{-3}+c$$$$=\frac{-1}{(1+tan\alpha )}+\frac{1}{(1+2tan\alpha +{tan}^2\alpha )}-\frac{2}{3{(1+tan\alpha )}^3}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com