Math, asked by BrainlyWarrior, 1 year ago

Solve :

\int \dfrac{Sinx}{Sin^{3}x + Cos^{3}x} . dx

Answers

Answered by Swarup1998
24
\underline{\text{Solution :}}

\star

\mathrm{Now,\:\frac{1}{z^{3}+1}=\frac{1}{(z+1)(z^{2}-z+1)}}

\to \mathrm{\frac{1}{(z+1)(z^{2}-z+1)}=\frac{A}{z+1}+\frac{Bz+C}{z^{2}-z+1}}

\to \mathrm{1=A(z^{2}-z+1)+(Bz+C)(z+1)}

\to \mathrm{1=Az^{2}-Az+A+Bz^{2}+Bz+Cz+C}

\to \mathrm{(A+B)z^{2}+(-A+B+C)z+(A+C)=1}

\text{Equating both sides, we get}

\mathrm{A+B=0.....(i)}

\mathrm{-A+B+C=0.....(ii)}

\mathrm{A+C=1.....(iii)}

\mathrm{(i)+(iii)-(ii)}\implies

\mathrm{A+B+A+C+A-B-C=1+0-0}

\to \mathrm{3A=1}

\to \mathrm{A=\frac{1}{3}}

\mathrm{Then,\:B=-A=-\frac{1}{3}}

\mathrm{and\:C=1-\frac{1}{3}=\frac{2}{3}}

\mathrm{Now,\:\int \frac{dz}{z^{3}+1}}

\mathrm{=\frac{1}{3}\int \frac{dz}{z+1}+\frac{1}{3}\int \frac{-z+2}{z^{2}-z+1}dz}

\mathrm{=\frac{1}{3}\int \frac{dz}{z+1}-\frac{1}{6}\int \frac{(2z-1)+1}{z^{2}-z+1}dz+2\int \frac{dz}{z^{2}-z+1}}

\mathrm{=\frac{1}{3}\int \frac{dz}{z+1}-\frac{1}{6}\int\frac{(2z-1)dz}{z^{2}-z+1}-\frac{1}{6}\int \frac{dz}{z^{2}-z+1}+2\int\frac{dz}{z^{2}-z+1}}

\mathrm{=\frac{1}{3}\int \frac{dz}{z+1}-\frac{1}{6}\int\frac{(2z-1)dz}{z^{2}-z+1}+\frac{11}{6}\int\frac{dz}{z^{2}-z+1}}

\star\star

\mathrm{Now,\:\int\frac{dz}{z^{2}-z+1}}

\mathrm{=\int\frac{dz}{(z-\frac{1}{2})^{2}-(\frac{\sqrt{3}}{2})^{2}}}

\mathrm{=\frac{1}{\frac{\sqrt{3}}{2}}tan^{-1}\frac{z}{\frac{\sqrt{3}}{2}}}

\mathrm{=\frac{2}{\sqrt{3}}tan^{-1}\frac{2z}{\sqrt{3}}}

\star\star\star

\mathrm{Now,\:\int\frac{sinx}{sin^{3}x+cos^{3}x}dx}

\mathrm{=\int \frac{tanx\:sec^{2}x}{tan^{3}x+1}dx}

\boxed{\mathrm{Let,\:tanx=z\to sec^{2}x\:dx=dz}}

\mathrm{=\int \frac{z\:dz}{z^{3}+1}}

\mathrm{=\int \frac{(z+1)-1}{z^{3}+1}dz}

\mathrm{=\int \frac{z+1}{(z+1)(z^{2}-z+1)}dz-\int \frac{dz}{z^{3}+1}}

\mathrm{=\int \frac{dz}{z^{2}-z+1}-\frac{dz}{z^{3}+1}}

\mathrm{=\int \frac{dz}{z^{2}-z+1}-\bigg(\frac{1}{3}\int \frac{dz}{z+1}-\frac{1}{6}\int\frac{(2z-1)dz}{z^{2}-z+1}+\frac{11}{6}\int\frac{dz}{z^{2}-z+1}\bigg)}

\mathrm{=\int \frac{dz}{z^{2}-z+1}-\frac{1}{3}\int \frac{dz}{z+1}+\frac{1}{6}\int\frac{(2z-1)dz}{z^{2}-z+1}-\frac{11}{6}\int\frac{dz}{z^{2}-z+1}}

\mathrm{=-\frac{5}{6}\int \frac{dz}{z^{2}-z+1}-\frac{1}{3}\int \frac{dz}{z+1}+\frac{1}{6}\int\frac{(2z-1)dz}{z^{2}-z+1}}

\mathrm{=-\frac{5}{6}\frac{2}{\sqrt{3}}tan^{-1}\frac{2z}{\sqrt{3}}-\frac{1}{3}log|z+1|+\frac{1}{6}log|z^{2}-z+1|+C}

\text{where C is integral constant}

\mathrm{=-\frac{5}{3\sqrt{3}}tan^{-1}\frac{2tanx}{\sqrt{3}}-\frac{1}{3}log|tanx+1|+\frac{1}{6}log|tan^{2}x-tanx+1|+C}

\to \boxed{\mathrm{\int\frac{sinx}{sin^{3}x+cos^{3}x}dx=-\frac{5}{3\sqrt{3}}tan^{-1}\frac{2tanx}{\sqrt{3}}-\frac{1}{3}log|tanx+1|+\frac{1}{6}log|tan^{2}x-tanx+1|+C}}

\text{which is the required integral.}

Swarup1998: Is it correct? @BrainlyWarrior
BrainlyWarrior: I'm not sure about this
Swarup1998: Calculations wrong in some places. Let me correct.
Swarup1998: Hey! It is correct. :)
BrainlyWarrior: okay!
Answered by brainlymasterking
0

\int  \dfrac{Sin \: x}{Sin^{3}x + Cos^{3}x} . dx \\  \int \frac{sin \: x}{( \sin \: x + cos \: x)(sin ^{2} x - sin \: x \: cos \: x + cos^{2 \: x}  } [:: a ^{2}  + b ^{3}  = (a + b)(a² – ab + b² )] = \\  \int \frac{sin \: x}{(sin \: x + cos \: x)(1 - sin \: x \: cos \: x} dx

Half solution will be continued in image explanation

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