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Question Number 144064 by 0731619 last updated on 21/Jun/21
Answered by MJS_new last updated on 21/Jun/21
simply let t=tan x ⇒ answer is −(x/b^2 )+((√(a^2 +b^2 ))/(ab^2 ))arctan ((atan x)/( (√(a^2 +b^2 )))) +C
$$\mathrm{simply}\:\mathrm{let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is} \\ $$$$−\frac{{x}}{{b}^{\mathrm{2}} }+\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{ab}^{\mathrm{2}} }\mathrm{arctan}\:\frac{{a}\mathrm{tan}\:{x}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:+{C} \\ $$
Commented by 0731619 last updated on 21/Jun/21
prove this
$${prove}\:{this} \\ $$
Commented by MJS_new last updated on 21/Jun/21
easy to prove (d/dx)[−(x/b^2 )]=−(1/b^2 ) (d/dx)[((√(a^2 +b^2 ))/(ab^2 ))arctan ((atan x)/( (√(a^2 +b^2 ))))]= =((√(a^2 +b^2 ))/(ab^2 ))×((a(√(a^2 +b^2 )))/(a^2 +b^2 +a^2 tan^2 x))×(1/(cos^2 x))= =((a^2 +b^2 )/(b^2 (a^2 +b^2 cos^2 x))) ((a^2 +b^2 )/(b^2 (a^2 +b^2 cos^2 x)))−(1/b^2 )=((sin^2 x)/(a^2 +b^2 cos^2 x)) q.e.d.
$$\mathrm{easy}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\frac{{d}}{{dx}}\left[−\frac{{x}}{{b}^{\mathrm{2}} }\right]=−\frac{\mathrm{1}}{{b}^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dx}}\left[\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{ab}^{\mathrm{2}} }\mathrm{arctan}\:\frac{{a}\mathrm{tan}\:{x}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right]= \\ $$$$=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{ab}^{\mathrm{2}} }×\frac{{a}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:{x}}×\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}= \\ $$$$=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}\right)} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}\right)}−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }=\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}}\:{q}.{e}.{d}. \\ $$
Commented by Dwaipayan Shikari last updated on 21/Jun/21
Q.E.D
$${Q}.{E}.{D} \\ $$
Answered by mathmax by abdo last updated on 21/Jun/21
Φ=∫ ((sin^2 x)/(b^2 cos^2 x +a^2 ))dx=∫ ((sin^2 xdx)/(b^2 (cos^2 x +(a^2 /b^2 ))))=(1/b^2 )∫ ((sin^2 x)/(λ +cos^2 x))dx with λ=(a^2 /b^2 )>0 f(λ)=∫ ((sin^2 x)/(λ+cos^2 x))dx =∫ (((1−cos(2x))/2)/(λ+((1+cos(2x))/2)))dx =∫ ((1−cos(2x))/(2λ+1+cos(2x)))dx =_(2x=t) (1/2) ∫ ((1−cost)/(2λ+1+cost))dt =_(tan((t/2))=y) (1/2)∫ ((1−((1−y^2 )/(1+y^2 )))/(2λ+1+((1−y^2 )/(1+y^2 ))))×((2dy)/(1+y^2 )) =∫ ((2y^2 )/((2λ+1)+(2λ+1)y^2 +1−y^2 )))(dy/((1+y^2 ))) =2∫ (y^2 /((1+y^2 )(2λ+2+2λy^2 )))dy=∫ (y^2 /((y^2 +1)(λ+1+y^2 )))dy F(y)=(y^2 /((y^2 +1)(y^2 +λ+1)))=(1/λ)((y^2 /(y^2 +1))−(y^2 /(y^2 +λ+1))) =(1/λ){((y^2 +1−1)/(y^2 +1))−((y^2 +λ+1−λ−1)/(y^2 +λ+1))} =(1/λ){−(1/(y^2 +1))+((λ+1)/(y^2 +λ+1))}
$$\Phi=\int\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \mathrm{x}\:+\mathrm{a}^{\mathrm{2}} }\mathrm{dx}=\int\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{xdx}}{\mathrm{b}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\int\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\lambda\:+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{with}\:\lambda=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }>\mathrm{0} \\ $$$$\mathrm{f}\left(\lambda\right)=\int\:\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\lambda+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:=\int\:\:\frac{\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}{\lambda+\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx} \\ $$$$=\int\:\:\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}\lambda+\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}\mathrm{dx}\:=_{\mathrm{2x}=\mathrm{t}} \:\frac{\mathrm{1}}{\mathrm{2}}\:\:\int\:\:\frac{\mathrm{1}−\mathrm{cost}}{\mathrm{2}\lambda+\mathrm{1}+\mathrm{cost}}\mathrm{dt} \\ $$$$=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{y}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{1}−\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}{\mathrm{2}\lambda+\mathrm{1}+\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}×\frac{\mathrm{2dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} } \\ $$$$=\int\:\:\:\frac{\mathrm{2y}^{\mathrm{2}} }{\left.\left(\mathrm{2}\lambda+\mathrm{1}\right)+\left(\mathrm{2}\lambda+\mathrm{1}\right)\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)}\frac{\mathrm{dy}}{\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}\int\:\:\:\frac{\mathrm{y}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{2}\lambda+\mathrm{2}+\mathrm{2}\lambda\mathrm{y}^{\mathrm{2}} \right)}\mathrm{dy}=\int\:\frac{\mathrm{y}^{\mathrm{2}} }{\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)\left(\lambda+\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}\mathrm{dy} \\ $$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{\mathrm{y}^{\mathrm{2}} }{\left(\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{y}^{\mathrm{2}} +\lambda+\mathrm{1}\right)}=\frac{\mathrm{1}}{\lambda}\left(\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} \:+\lambda+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\lambda}\left\{\frac{\mathrm{y}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{y}^{\mathrm{2}} \:+\lambda+\mathrm{1}−\lambda−\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\lambda+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\lambda}\left\{−\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}+\frac{\lambda+\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \:+\lambda+\mathrm{1}}\right\} \\ $$
Commented by mathmax by abdo last updated on 21/Jun/21
f(λ)=−(1/λ)∫ (dy/(y^2 +1))+((λ+1)/λ)∫ (dy/(y^2 +λ+1))(→y=(√(λ+1))z) =−(1/λ)arctany +((λ+1)/λ)∫ (((√(λ+1))dz)/((λ+1)(1+z^2 ))) =−(1/λ)arctany+((√(λ+1))/λ) arctan((y/( (√(λ+1)))))+k =−(x/λ) +((√(λ+1))/λ)arctan((1/( (√(λ+1))))tanx) +K Φ=(1/b^2 )(−(x/(a^2 /b^2 ))+(b^2 /a^2 )(√((a^2 /b^2 )+1))arctan((1/( (√((a^2 /b^2 )+1))))tanx)+K =−(x/a^2 )+(1/a^2 )(√((a^2 +b^2 )/b^2 ))arctan(((∣b∣)/( (√(a^2 +b^2 ))))tanx) +K
$$\mathrm{f}\left(\lambda\right)=−\frac{\mathrm{1}}{\lambda}\int\:\frac{\mathrm{dy}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}+\frac{\lambda+\mathrm{1}}{\lambda}\int\:\frac{\mathrm{dy}}{\mathrm{y}^{\mathrm{2}} \:+\lambda+\mathrm{1}}\left(\rightarrow\mathrm{y}=\sqrt{\lambda+\mathrm{1}}\mathrm{z}\right) \\ $$$$=−\frac{\mathrm{1}}{\lambda}\mathrm{arctany}\:+\frac{\lambda+\mathrm{1}}{\lambda}\int\:\frac{\sqrt{\lambda+\mathrm{1}}\mathrm{dz}}{\left(\lambda+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)} \\ $$$$=−\frac{\mathrm{1}}{\lambda}\mathrm{arctany}+\frac{\sqrt{\lambda+\mathrm{1}}}{\lambda}\:\mathrm{arctan}\left(\frac{\mathrm{y}}{\:\sqrt{\lambda+\mathrm{1}}}\right)+\mathrm{k} \\ $$$$=−\frac{\mathrm{x}}{\lambda}\:+\frac{\sqrt{\lambda+\mathrm{1}}}{\lambda}\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\lambda+\mathrm{1}}}\mathrm{tanx}\right)\:+\mathrm{K} \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\left(−\frac{\mathrm{x}}{\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\sqrt{\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }+\mathrm{1}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }+\mathrm{1}}}\mathrm{tanx}\right)+\mathrm{K}\right. \\ $$$$=−\frac{\mathrm{x}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\sqrt{\frac{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }}\mathrm{arctan}\left(\frac{\mid\mathrm{b}\mid}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}\mathrm{tanx}\right)\:+\mathrm{K} \\ $$

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