Menu Close

Question-193526

Tinku Tara 0 Comments
Pin




Question Number 193526 by SaRahAli last updated on 16/Jun/23
Answered by MM42 last updated on 16/Jun/23
I=∫ ((sinx×cosx)/(sinx+cos^2 x)) dx=−∫ ((sinx×cosx)/(sinx−sinx−1)) dx let sinx=u⇒cosxdx=du ⇒I=−∫ ((udu)/(u^2 −u−1)) du =∫ (A/(u−u_1 )) du +∫ (B/(u−u_2 )) du the solution is very simple
$${I}=\int\:\frac{{sinx}×{cosx}}{{sinx}+{cos}^{\mathrm{2}} {x}}\:{dx}=−\int\:\frac{{sinx}×{cosx}}{{sinx}−{sinx}−\mathrm{1}}\:{dx} \\ $$$${let}\:\:\:{sinx}={u}\Rightarrow{cosxdx}={du} \\ $$$$\Rightarrow{I}=−\int\:\frac{{udu}}{{u}^{\mathrm{2}} −{u}−\mathrm{1}}\:{du}\:=\int\:\frac{{A}}{{u}−{u}_{\mathrm{1}} }\:{du}\:+\int\:\frac{{B}}{{u}−{u}_{\mathrm{2}} }\:{du}\: \\ $$$${the}\:{solution}\:{is}\:{very}\:{simple} \\ $$
Answered by Subhi last updated on 16/Jun/23
∫((sin(x))/(tan(x)+cos(x)))dx ⇛ ∫((sin(x))/(((sin(x))/(cos(x)))+cos(x))) ∫((sin(x).cos(x))/(sin(x)+1−sin^2 (x))) ⇛ sin^2 (x)=u ⇛ 2sin(x)cos(x)dx=du ((−1)/2)∫(du/(u−(√u)−1)) ⇛ (√u)=v ⇛ (1/(2(√u)))du=dv ((−1)/2)∫((2v )/(v^2 −v−1))dv ⇛ −∫(v/(v^2 −v−1)) ⇛ −∫(a/(v−((1+(√5))/2)))+(b/(v−((1−(√5))/2))) av−a.((1−(√5))/2)+bv−b.((1+(√5))/2) a+b = 1 ↬ −((1−(√5))/2)a − ((1+(√5))/2)b=0 (√5)a = ((1+(√5))/2) ⇛ a = ((5+(√5))/(10)) ⇛ b = ((5−(√5))/(10)) −((5+(√5))/(10))∫(1/(v−((1+(√5))/2))) −((5−(√5))/(10))∫(1/(v−((1−(√5))/2))) −((5+(√5))/(10)).ln∣sin(x)−((1+(√5))/2)∣−((5−(√5))/(10)).ln∣sin(x)−((1−(√5))/2)∣
$$\int\frac{{sin}\left({x}\right)}{{tan}\left({x}\right)+{cos}\left({x}\right)}{dx}\:\:\Rrightarrow\:\int\frac{{sin}\left({x}\right)}{\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}+{cos}\left({x}\right)} \\ $$$$\int\frac{{sin}\left({x}\right).{cos}\left({x}\right)}{{sin}\left({x}\right)+\mathrm{1}−{sin}^{\mathrm{2}} \left({x}\right)}\:\Rrightarrow\:{sin}^{\mathrm{2}} \left({x}\right)={u}\:\Rrightarrow\:\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right){dx}={du} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}−\sqrt{{u}}−\mathrm{1}}\:\:\Rrightarrow\:\sqrt{{u}}={v}\:\Rrightarrow\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{u}}}{du}={dv} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{v}\:}{{v}^{\mathrm{2}} −{v}−\mathrm{1}}{dv}\:\Rrightarrow\:−\int\frac{{v}}{{v}^{\mathrm{2}} −{v}−\mathrm{1}}\:\Rrightarrow\:−\int\frac{{a}}{{v}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}+\frac{{b}}{{v}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$${av}−{a}.\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}+{bv}−{b}.\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${a}+{b}\:=\:\mathrm{1}\:\:\:\:\:\:\looparrowright\:\:−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{a}\:−\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{b}=\mathrm{0} \\ $$$$\sqrt{\mathrm{5}}{a}\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\Rrightarrow\:{a}\:=\:\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{10}}\:\:\Rrightarrow\:{b}\:=\:\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{10}}\:\: \\ $$$$−\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{10}}\int\frac{\mathrm{1}}{{v}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:−\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{10}}\int\frac{\mathrm{1}}{{v}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$$−\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{10}}.{ln}\mid{sin}\left({x}\right)−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\mid−\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{10}}.{ln}\mid{sin}\left({x}\right)−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\mid \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *