Substitute u = x 2 u=x^2 u = x 2 d y d u + 2 y sin 2 u ln tan u = 2 2 sin ( u − π / 4 ) sin 2 u ln tan u \frac{dy}{du}+\frac{2y}{\sin2u\ln\tan u}=\frac{2\sqrt2\sin(u-\pi/4)}{\sin2u\ln\tan u} d u d y + s i n 2 u l n t a n u 2 y = s i n 2 u l n t a n u 2 2 s i n ( u − π /4 )
Integrating factor is ln tan u \ln\tan u ln tan u 2 2 sin ( u − π / 4 ) sin 2 u = sec u − cosec u \frac{2\sqrt2\sin(u-\pi/4)}{\sin2u}=\sec u-\operatorname{cosec}u s i n 2 u 2 2 s i n ( u − π /4 ) = sec u − cosec u
So y ln tan u = ln ∣ sec u + tan u ∣ − ln ∣ tan u 2 ∣ + C y\ln\tan u=\ln|\sec u+\tan u|-\ln\left|\tan\frac u2\right|+C y ln tan u = ln ∣ sec u + tan u ∣ − ln tan 2 u + C
Constant. At u = π 6 u=\frac\pi6 u = 6 π y = 1 y=1 y = 1 C = ln ( 2 − 3 ) − ln 3 C=\ln(2-\sqrt3)-\ln3 C = ln ( 2 − 3 ) − ln 3
Evaluate at u = π 3 u=\frac\pi3 u = 3 π ln 3 2 y = ln ( 2 + 3 ) + ln 3 2 + ln ( 2 − 3 ) − ln 3 = − ln 3 2 \frac{\ln3}2\,y=\ln(2+\sqrt3)+\frac{\ln3}2+\ln(2-\sqrt3)-\ln3=-\frac{\ln3}2 2 l n 3 y = ln ( 2 + 3 ) + 2 l n 3 + ln ( 2 − 3 ) − ln 3 = − 2 l n 3 y = − 1 y=-1 y = − 1
Answer: 1