Question Number 169354 by Shrinava last updated on 29/Apr/22
$$\mathrm{1}.\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}\:+\:\mathrm{x}} \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{6}\:-\:\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cosx}}{\mathrm{1}\:+\:\mathrm{sinx}}\:\mathrm{dx} \\ $$
Answered by greougoury555 last updated on 29/Apr/22
![(3) ∫_0 ^( (π/2)) ((d(sin x+1))/(1+sin x)) = [ ln (1+sin x) ]_0 ^(π/2) = ln 2 (2) ∫_0 ^( 1) (6−x^2 )dx= [6x−(1/3)x^3 ]_0 ^1 = 6−(1/3)=((17)/3)](https://www.tinkutara.com/question/Q169355.png)
$$\left(\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{sin}\:{x}+\mathrm{1}\right)}{\mathrm{1}+\mathrm{sin}\:{x}}\:=\:\left[\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{ln}\:\mathrm{2}\: \\ $$$$\left(\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{6}−{x}^{\mathrm{2}} \right){dx}=\:\left[\mathrm{6}{x}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \:\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{6}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{17}}{\mathrm{3}} \\ $$