Question Number 32675 by Cheyboy last updated on 31/Mar/18
Commented by caravan msup abdo. last updated on 31/Mar/18
![we have 1+cos(4x)=2cos^2 (2x)⇒ I= ∫_0 ^(π/2) (√(1+cos(4x))) dx = (√2) ∫_0 ^(π/2) ∣cos(2x)∣dx =(√2) ∫_0 ^(π/4) cos(2x)dx −(√2) ∫_(π/4) ^(π/2) cos(2x)dx =((√2)/2) [ sin(2x)]_0 ^(π/4) −((√2)/2) [sin(2x)]_(π/4) ^(π/2) =((√2)/2) −((√2)/2)(−1) =(√2) .](https://www.tinkutara.com/question/Q32700.png)
$${we}\:{have}\:\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\Rightarrow \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}\:{dx} \\ $$$$=\:\sqrt{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mid{cos}\left(\mathrm{2}{x}\right)\mid{dx} \\ $$$$=\sqrt{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{x}\right){dx}\:−\sqrt{\mathrm{2}}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\left(\mathrm{2}{x}\right){dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left[\:{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left[{sin}\left(\mathrm{2}{x}\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(−\mathrm{1}\right)\:=\sqrt{\mathrm{2}}\:\:. \\ $$
Answered by MJS last updated on 31/Mar/18
$$\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2cos}^{\mathrm{2}} \:{x}−\mathrm{1} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \:\mathrm{2}{x}−\mathrm{1}=\mathrm{8cos}^{\mathrm{4}} \:{x}−\mathrm{8cos}^{\mathrm{2}} \:{x}+\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{1}+\mathrm{cos}\:\mathrm{4}{x}=\mathrm{2}\left(\mathrm{4cos}^{\mathrm{4}} \:{x}−\mathrm{4cos}^{\mathrm{2}} \:{x}+\mathrm{1}\right)= \\ $$$$=\mathrm{2}\left(\mathrm{2cos}^{\mathrm{2}} \:{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2cos}^{\mathrm{2}} \:\mathrm{2}{x} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\sqrt{\mathrm{2}}\mid\mathrm{cos}\:\mathrm{2}{x}\mid{dx}=\mathrm{2}\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:\mathrm{2}{x}\:{dx}= \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\mid}}=\sqrt{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\mid}}=\sqrt{\mathrm{2}} \\ $$
Commented by Cheyboy last updated on 31/Mar/18
$${Thank}\:{you}\:{sir}\:{bt}\:{the}\:{upper} \\ $$$${limit}\:{ir}\:\pi/\mathrm{2} \\ $$
Commented by MJS last updated on 31/Mar/18
![I changed the upper limit because I used cos 2x instead of ∣cos 2x∣. cos 2x is ≥0 in [0;(π/4)] but ≤0 in [(π/4);(π/2)], so (√2)∫_0 ^(π/2) cos 2x dx=0 but ∫_0 ^(π/4) cos 2x dx=−∫_(π/4) ^(π/2) cos 2x dx so (√2)∫_0 ^(π/2) ∣cos 2x∣dx=2(√2)∫_0 ^(π/4) cos 2x dx](https://www.tinkutara.com/question/Q32687.png)
$$\mathrm{I}\:\mathrm{changed}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{limit} \\ $$$$\mathrm{because}\:\mathrm{I}\:\mathrm{used}\:\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{instead} \\ $$$$\mathrm{of}\:\mid\mathrm{cos}\:\mathrm{2}{x}\mid.\:\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{is}\:\geqslant\mathrm{0}\:\mathrm{in}\:\left[\mathrm{0};\frac{\pi}{\mathrm{4}}\right] \\ $$$$\mathrm{but}\:\leqslant\mathrm{0}\:\mathrm{in}\:\left[\frac{\pi}{\mathrm{4}};\frac{\pi}{\mathrm{2}}\right],\:\mathrm{so}\:\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{cos}\:\mathrm{2}{x}\:{dx}=\mathrm{0} \\ $$$$\mathrm{but}\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:\mathrm{2}{x}\:{dx}=−\underset{\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{cos}\:\mathrm{2}{x}\:{dx}\:\mathrm{so} \\ $$$$\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mid\mathrm{cos}\:\mathrm{2}{x}\mid{dx}=\mathrm{2}\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:\mathrm{2}{x}\:{dx} \\ $$$$ \\ $$
Commented by Cheyboy last updated on 31/Mar/18
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}}\:\boldsymbol{{thatz}}\:\boldsymbol{{well}}\:\boldsymbol{{understood}} \\ $$