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Question Number 127870 by Eric002 last updated on 02/Jan/21

2021 HAPPY NEW Year 1)∫((x^3 +3x+2)/((x^2 +1)^2 (x+1)))dx 2)∫((2cos(x)−sin(x))/(3sin(x)+5cos(x)))dx 3)∫((tan(2x))/( (√(sin^6 (x)+cos^6 (x)))))dx 4)∫x(√((1−x^2 )/(1+x^2 ))) dx

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2021} \\ $$$${HAPPY}\:{NEW}\:{Year} \\ $$$$\left.\mathrm{1}\right)\int\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\int\frac{\mathrm{2}{cos}\left({x}\right)−{sin}\left({x}\right)}{\mathrm{3}{sin}\left({x}\right)+\mathrm{5}{cos}\left({x}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\int\frac{{tan}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}^{\mathrm{6}} \left({x}\right)+{cos}^{\mathrm{6}} \left({x}\right)}}{dx} \\ $$$$ \\ $$$$\left.\mathrm{4}\right)\int{x}\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Jan/21

∫x(√((1−x^2 )/(1+x^2 ))) dx x^2 =u⇒2x=(du/dx) =(1/2)∫(√((1−u)/(1+u))) =(1/2)∫(1/( (√(1−u^2 ))))−(u/(2(√(1−u^2 ))))du =(1/2)sin^(−1) (x^2 )+(1/2)(√(1−x^4 ))

$$\int{x}\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u}\Rightarrow\mathrm{2}{x}=\frac{{du}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}−\frac{{u}}{\mathrm{2}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }\:\: \\ $$

Commented by Eric002 last updated on 02/Jan/21

thank you

$${thank}\:{you} \\ $$

Answered by liberty last updated on 03/Jan/21

(2)∫ ((−sin x+2cos x)/(3sin x+5cos x)) dx = ((−sin x+2cos x)/(3sin x+5cos x)) = P(((3sin x+5cos x )/(3sin x+5cos x)))+Q(((3cos x−5sin x)/(3sin x+5cos x))) ⇒−sin x+2cos x=(3P−5Q)sin x+(5P+3Q)cos x (((3 −5)),((5 3)) ) ((P),(Q) ) = (((−1)),(( 2)) ) { ((P=( determinant (((−1 −5)),(( 2 3)))/(34))=(7/(34)))),((Q=( determinant (((3 −1)),((5 2)))/(34))=((11)/(34)))) :} then ∫ ((−sin x+2cos x)/(3sin x+5cos x)) dx = ((7x)/(34))+((11)/(34)) ℓn ∣3sin x+5cos x ∣ + C

$$\left(\mathrm{2}\right)\int\:\frac{−\mathrm{sin}\:\mathrm{x}+\mathrm{2cos}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{x}+\mathrm{5cos}\:\mathrm{x}}\:\mathrm{dx}\:= \\ $$$$\:\frac{−\mathrm{sin}\:\mathrm{x}+\mathrm{2cos}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{x}+\mathrm{5cos}\:\mathrm{x}}\:=\:\mathrm{P}\left(\frac{\mathrm{3sin}\:\mathrm{x}+\mathrm{5cos}\:\mathrm{x}\:}{\mathrm{3sin}\:\mathrm{x}+\mathrm{5cos}\:\mathrm{x}}\right)+\mathrm{Q}\left(\frac{\mathrm{3cos}\:\mathrm{x}−\mathrm{5sin}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{x}+\mathrm{5cos}\:\mathrm{x}}\right) \\ $$$$\Rightarrow−\mathrm{sin}\:\mathrm{x}+\mathrm{2cos}\:\mathrm{x}=\left(\mathrm{3P}−\mathrm{5Q}\right)\mathrm{sin}\:\mathrm{x}+\left(\mathrm{5P}+\mathrm{3Q}\right)\mathrm{cos}\:\mathrm{x} \\ $$$$\:\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:−\mathrm{5}}\\{\mathrm{5}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{P}}\\{\mathrm{Q}}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{1}}\\{\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{cases}{\mathrm{P}=\frac{\begin{vmatrix}{−\mathrm{1}\:\:\:\:−\mathrm{5}}\\{\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{3}}\end{vmatrix}}{\mathrm{34}}=\frac{\mathrm{7}}{\mathrm{34}}}\\{\mathrm{Q}=\frac{\begin{vmatrix}{\mathrm{3}\:\:\:\:−\mathrm{1}}\\{\mathrm{5}\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}}{\mathrm{34}}=\frac{\mathrm{11}}{\mathrm{34}}}\end{cases}\: \\ $$$$\mathrm{then}\:\int\:\frac{−\mathrm{sin}\:\mathrm{x}+\mathrm{2cos}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{x}+\mathrm{5cos}\:\mathrm{x}}\:\mathrm{dx}\:=\:\frac{\mathrm{7x}}{\mathrm{34}}+\frac{\mathrm{11}}{\mathrm{34}}\:\ell\mathrm{n}\:\mid\mathrm{3sin}\:\mathrm{x}+\mathrm{5cos}\:\mathrm{x}\:\mid\:+\:\mathrm{C} \\ $$

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