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Question Number 148452 by mathdanisur last updated on 28/Jul/21

$$\underset{1}{\stackrel{4}{\int }}2{sin}^{2}xdx+\underset{1}{\stackrel{4}{\int }}(1+cos2x)dx=?$$

Answered by puissant last updated on 28/Jul/21

$$={\int }_1^{4}2\left(\frac{1-\cos 2\mathrm{x}}{2}\right)+(1+\cos 2\mathrm{x})\mathrm{dx}$$$$={\int }_1^4(1-\cos 2\mathrm{x}+1+\cos 2\mathrm{x})\mathrm{dx}$$$$={\int }_1^{4}2\mathrm{dx}=2{\left[\mathrm{x}\right]}_1^4=2(4-1)$$$$\mathrm{I}=6...$$

Commented by mathdanisur last updated on 28/Jul/21

$$ThankyouSer$$

Answered by Ar Brandon last updated on 28/Jul/21

$$1+\cos 2\mathrm{x}=({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x})+({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x})={2\cos }^2\mathrm{x}$$$$\mathrm{I}={\int }_1^4{2\sin }^2\mathrm{xdx}+{\int }_1^4{2\cos }^2\mathrm{xdx}$$$$=2{\int }_1^4({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})\mathrm{dx}=2{\int }_1^4\mathrm{dx}=2(4-1)=6$$

Commented by mathdanisur last updated on 28/Jul/21

$$ThankyouSer$$

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