Đặt πt14=u⇒u∈[0;12π7)πt14=u⇒u∈[0;12π7) khi đó ta có

h=2sin(3u)(1−4sin2u)+12⇔h=2(3sinu−4sin3u)(1−4sin2u)+12h=2sin⁡(3u)(1−4sin2u)+12⇔h=2(3sinu−4sin3u)(1−4sin2u)+12

Đặt

v=sinu⇒h(v)=2(3t−4t3)(1−4t2)+12=6t−24t3−8t3+32t5+12=32t5−32t3+6t−12v=sin⁡u⇒h(v)=2(3t−4t3)(1−4t2)+12=6t−24t3−8t3+32t5+12=32t5−32t3+6t−12

Xét u∈[0;π2]⇒v∈[0;1]