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Question Number 138691 by liberty last updated on 16/Apr/21

∫ cos 2x (√(1+sin^2 x)) dx =?

$$\int\:\mathrm{cos}\:\mathrm{2}{x}\:\sqrt{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\:=? \\ $$

Answered by mathmax by abdo last updated on 17/Apr/21

I=∫ cos(2x)(√(1+sin^2 x))dx ⇒I=∫ cos(2x)(√(1+((1−cos(2x))/2)))dx =∫ cos(2x)(√((3−cos(2x))/2))dx =(1/( (√2)))∫cos(2x)(√(3−cos(2x)))dx =_(2x=t) (1/( (√2)))∫ cost(√(3−cost))(dt/2) =(1/(2(√2)))∫ cost(√(3−cost))dt =_(cost=y) (1/(2(√2)))∫ y(√(3−y))(−(dy/( (√(1−y^2 ))))) =−(1/(2(√2)))∫ (y/( (√(1−y^2 ))))(√(3−y))dy by parts f^′ =(y/( (√(1−y^2 )))) and g=(√(3−y)) ⇒ ∫ (y/( (√(1−y^2 ))))(√(3−y))dy =−(√(1−y^2 ))(√(3−y))−∫−(√(1−y^2 ))(((−1)/(2(√(3−y)))))dy =−(√(3−y^2 ))(√(1−y^2 ))−∫((√(1−y^2 ))/(2(√(3−y))))dy changement (√(3−y))=u give 3−y =u^2 ⇒y =3+u^2 ⇒∫ ((√(1−y^2 ))/(2(√(3−y))))dy =∫ ((√(1−(3+u^2 )^2 ))/(2u))(2u)du =∫ (√(1−(u^2 +3)^2 ))du ....be continued...

$$\mathrm{I}=\int\:\mathrm{cos}\left(\mathrm{2x}\right)\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\Rightarrow\mathrm{I}=\int\:\mathrm{cos}\left(\mathrm{2x}\right)\sqrt{\mathrm{1}+\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx} \\ $$$$=\int\:\mathrm{cos}\left(\mathrm{2x}\right)\sqrt{\frac{\mathrm{3}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\mathrm{cos}\left(\mathrm{2x}\right)\sqrt{\mathrm{3}−\mathrm{cos}\left(\mathrm{2x}\right)}\mathrm{dx} \\ $$$$=_{\mathrm{2x}=\mathrm{t}} \:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:\mathrm{cost}\sqrt{\mathrm{3}−\mathrm{cost}}\frac{\mathrm{dt}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\mathrm{cost}\sqrt{\mathrm{3}−\mathrm{cost}}\mathrm{dt} \\ $$$$=_{\mathrm{cost}=\mathrm{y}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\mathrm{y}\sqrt{\mathrm{3}−\mathrm{y}}\left(−\frac{\mathrm{dy}}{\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\:\frac{\mathrm{y}}{\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}\sqrt{\mathrm{3}−\mathrm{y}}\mathrm{dy} \\ $$$$\mathrm{by}\:\mathrm{parts}\:\:\mathrm{f}^{'} \:=\frac{\mathrm{y}}{\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}\:\mathrm{and}\:\mathrm{g}=\sqrt{\mathrm{3}−\mathrm{y}}\:\Rightarrow \\ $$$$\int\:\frac{\mathrm{y}}{\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}\sqrt{\mathrm{3}−\mathrm{y}}\mathrm{dy}\:=−\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\sqrt{\mathrm{3}−\mathrm{y}}−\int−\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\left(\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}−\mathrm{y}}}\right)\mathrm{dy} \\ $$$$=−\sqrt{\mathrm{3}−\mathrm{y}^{\mathrm{2}} }\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }−\int\frac{\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{3}−\mathrm{y}}}\mathrm{dy}\:\:\mathrm{changement}\:\sqrt{\mathrm{3}−\mathrm{y}}=\mathrm{u}\:\mathrm{give} \\ $$$$\mathrm{3}−\mathrm{y}\:=\mathrm{u}^{\mathrm{2}} \:\Rightarrow\mathrm{y}\:=\mathrm{3}+\mathrm{u}^{\mathrm{2}} \:\Rightarrow\int\:\frac{\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{3}−\mathrm{y}}}\mathrm{dy} \\ $$$$=\int\:\frac{\sqrt{\mathrm{1}−\left(\mathrm{3}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2u}}\left(\mathrm{2u}\right)\mathrm{du}\:=\int\:\sqrt{\mathrm{1}−\left(\mathrm{u}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }\mathrm{du} \\ $$$$....\mathrm{be}\:\mathrm{continued}... \\ $$

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