Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 81217 by M±th+et£s last updated on 10/Feb/20

Answered by mind is power last updated on 10/Feb/20

$$iwillposteallsolutionlater!$$$$for2ndSommemistacks$$$$ifound$$$$f\left(x\right)=\int \frac{\sqrt{tan\left(2x\right)}}{sin\left(x\right)}dx=2\sqrt{2.tan\left(x\right)}.{}_2{F}_1(\frac{1}{2},\frac{1}{8},\frac{9}{8};{tan}^4\left(x\right))+c$$$$tan\left(2x\right)=\frac{2tan\left(x\right)}{1-{tan}^2\left(x\right)},\mathrm{\forall }x\in [0,\frac{\pi }{4}[$$$$sin\left(x\right)=\sqrt{{sin}^2\left(x\right)},=\sqrt{\frac{{tan}^2\left(x\right)}{1+{tan}^2\left(x\right)}}$$$$\int \frac{\sqrt{tan\left(2x\right)}}{sin\left(x\right)}dx=\int \sqrt{\frac{2tan\left(x\right)}{1-{tan}^2\left(x\right)}}.\sqrt{\frac{1+{tan}^2\left(x\right)}{{tan}^2\left(x\right)}}dx$$$$=\int \sqrt{\frac{2}{tan\left(x\right)}}.(1+{tan}^2\left(x\right)).\frac{dx}{\sqrt{1-{tan}^4\left(x\right)}}$$$$\frac{1}{\sqrt{1-x^2}}=\sum _{n⩾0}\frac{(2n!)}{2^{2n}.{(n!)}^2}{x}^{2n}$$$$\Rightarrow \frac{1}{\sqrt{1-{tan}^4\left(x\right)}}=\sum _{n⩾0}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}{tg}^{4n}\left(x\right)$$$$\int \sqrt{\frac{2}{tan\left(x\right)}}(1+{tan}^2\left(x\right)).\frac{dx}{\sqrt{1-{tan}^4\left(x\right)}}$$$$=\int \underset{n=0}{\sum ^{+\mathrm{\infty }}}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}.{tg}^{4n-\frac{1}{2}}\left(x\right)(1+{tg}^2\left(x\right))\sqrt{2}dx$$$$=\underset{n=0}{\sum ^{+\mathrm{\infty }}}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}\sqrt{2}\int {tg}^{4n-\frac{1}{2}}\left(x\right)(1+{tg}^2\left(x\right))dx$$$$=\underset{n=0}{\sum ^{+\mathrm{\infty }}}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}\sqrt{2}.\frac{{tg}^{4n+\frac{1}{2}}}{4n+\frac{1}{2}}+c=2\sqrt{2}\sum _{n⩾0}\frac{\left(2n\right)!{tg}^{4n+\frac{1}{2}}}{2^{2n}{(n!)}^2(8n+1)}+c$$$$$$$$=2\sqrt{2}.\sqrt{tg\left(x\right)}.\sum _{n⩾0}\frac{\left(2n\right)!!.{tg}^{4n}\left(x\right)}{2^{2n}{(n!)}^2(8n+1)}+c$$$$\left(2n\right)!=(2n-1)!!.2^n.n!$$$$f\left(x\right)=2\sqrt{2tan\left(x\right)}(1+\sum _{n⩾1}\frac{(2n-1)!!.2^{n}n!}{2^{2n}.{(n!)}^2(8n+1)}.{\left({tg}^4\left(x\right)\right)}^n)+c$$$$=2\sqrt{2tan\left(x\right)}(1+\sum _{n⩾1}\frac{(2n-1)!!}{2^n}.\frac{1}{8n+1}.\frac{{\left({tg}^4\left(x\right)\right)}^n}{n!})+c$$$$\frac{(2n-1)!!}{2^n}=\left(\frac{1}{2}\right)......(\frac{1}{2}+n-1)$$$$8n+1=8(n+\frac{9}{8}-1)\iff$$$$\frac{1}{8n+1}=\frac{\frac{1}{8}.......(\frac{1}{8}+n-1)}{\frac{9}{8}........(\frac{9}{8}+n-1)}$$$$\Rightarrow f\left(x\right)=2\sqrt{2tan\left(x\right)}(1+\sum _{n⩾1}\frac{\underset{k=1}{\prod ^n}(\frac{1}{2}+k-1)(\frac{1}{8}+k-1)}{\underset{k=1}{\prod ^n}(\frac{9}{8}+k-1)}.\frac{{\left({tg}^4\left(x\right)\right)}^n}{n!})+c$$$$1+\sum _{n⩾1}\frac{\left(a_n\right)\left(b_n\right)}{c_n}.\frac{x^n}{n!}={}_2{F}_1(a,b,c;x)$$$$f\left(x\right)=2\sqrt{2tan\left(x\right)}{}_2{F}_1(\frac{1}{2},\frac{1}{8},\frac{9}{8};{tg}^4\left(x\right))+c,ccanstante$$$$\int \frac{\sqrt{tan\left(2x\right)}}{sin\left(x\right)}dx=2\sqrt{2tan\left(x\right)}2F1(\frac{1}{2},\frac{1}{8},\frac{9}{8};{tg}^4\left(x\right))+c$$$$$$$$$$$$$$$$$$$$$$

Commented by M±th+et£s last updated on 10/Feb/20

$$greatsirgodblessyou$$

Commented by mind is power last updated on 10/Feb/20

$$thankyousirforposting$$$$sommhypergeometricFunction$$

Answered by M±th+et£s last updated on 11/Feb/20

Commented by mind is power last updated on 12/Feb/20

$$nice$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com