Integration Questions

Question Number 52900 by MJS last updated on 15/Jan/19

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}\:{x}\:\sqrt{\mathrm{sin}\:\mathrm{2}{x}}\:{dx}=? \\$$$$\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:{dx}=? \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sinx}\sqrt{{sin}\mathrm{2}{x}}\:{dx} \\$$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosx}\sqrt{{sin}\mathrm{2}{x}}\:{dx}\:\:\left[\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx}\right] \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cosx}+{sinx}\right)\sqrt{{sin}\mathrm{2}{x}}\:{dx} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {d}\left({sinx}−{cosx}\right)\sqrt{\mathrm{1}−\left(\mathrm{1}−{sin}\mathrm{2}{x}\right)}\:{dx} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {d}\left({sinx}−{cosx}\right)\sqrt{\mathrm{1}−\left({sinx}−{cosx}\right)^{\mathrm{2}} }\:{dx} \\$$$$\mid\frac{\left({sinx}−{cosx}\right)\sqrt{\mathrm{1}−\left({sinx}−{cosx}\right)^{\mathrm{2}} }}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{sinx}−{cosx}}{\mathrm{1}}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$=\left[\left\{\frac{\left(\mathrm{1}−\mathrm{0}\right)\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{0}\right)^{\mathrm{2}} }}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}}\right)\right\}−\left\{\frac{\mathrm{0}−\mathrm{1}\sqrt{\mathrm{1}−\mathrm{1}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{0}−\mathrm{1}}{\mathrm{1}}\right)\right\}\right] \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}−\left\{\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{−\pi}{\mathrm{2}}\right)\right\} \\$$$$=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}} \\$$$$\boldsymbol{{so}}\:\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}\:\:{sir}\:{pls}\:{check}... \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$$\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \:{cosx}\sqrt{{cos}\mathrm{2}{x}}\:{dx} \\$$$$\\$$$$\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\mathrm{2}×\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:\left[{here}\:{f}\left({x}\right)={evenfunction}\right] \\$$$${so}\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \:{cosx}\sqrt{{cos}\mathrm{2}{x}}\:{dx}=\mathrm{2}×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cosx}\sqrt{{cos}\mathrm{2}{x}}\:{dx} \\$$$${now}\:{using}\:{help}\:{of}\:{graph}... \\$$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$${so}\:\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \:{cosx}\sqrt{{cos}\mathrm{2}{x}}\:{dx}={area}\:{under}\:{the}\:{curve} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\frac{\mathrm{2}}{\mathrm{3}}×\frac{\pi}{\mathrm{2}}×\mathrm{1}=\frac{\pi}{\mathrm{3}} \\$$$${roughly}\:{it}\:{is}\:{a}\:{parabola}\:{sir}...{so}\:{area} \\$$$$\frac{\mathrm{2}}{\mathrm{3}}{ab} \\$$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

Answered by mr W last updated on 16/Jan/19

$$\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:{dx} \\$$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:{dx} \\$$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:{x}\sqrt{\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}}\:{dx} \\$$$$=\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\sqrt{\mathrm{1}−\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:{d}\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}\right)\:\:\:\:\:\:\:\:\left(\int\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\right) \\$$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left[\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}\right)+\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}\right)\sqrt{\mathrm{1}−\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\$$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\right) \\$$$$=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}} \\$$