Integration techniquesRandomize Question code: 2a634 Question 0803 Find the exact value of ∫034πsin6xcos6x dx.{\displaystyle \int_0^{\frac{3}{4} \pi} \sin 6 x \cos 6 x \, \mathrm{d}x.}∫043πsin6xcos6xdx. Hence find the exact value of ∫034π(sin6x+cos6x)2 dx.{\displaystyle \int_0^{\frac{3}{4} \pi} (\sin 6 x + \cos 6 x)^2 \, \mathrm{d}x.}∫043π(sin6x+cos6x)2dx. Answer ∫034πsin6xcos6x dx=112{\displaystyle \int_0^{\frac{3}{4} \pi} \sin 6 x \cos 6 x \, \mathrm{d}x = \frac{1}{12}}∫043πsin6xcos6xdx=121. ∫034π(sin6x+cos6x)2 dx=16+34π{\displaystyle \int_0^{\frac{3}{4} \pi} (\sin 6 x + \cos 6 x)^2 \, \mathrm{d}x = \frac{1}{6} + \frac{3}{4} \pi}∫043π(sin6x+cos6x)2dx=61+43π. Randomize Question code: 2a634
Find the exact value of ∫034πsin6xcos6x dx.{\displaystyle \int_0^{\frac{3}{4} \pi} \sin 6 x \cos 6 x \, \mathrm{d}x.}∫043πsin6xcos6xdx. Hence find the exact value of ∫034π(sin6x+cos6x)2 dx.{\displaystyle \int_0^{\frac{3}{4} \pi} (\sin 6 x + \cos 6 x)^2 \, \mathrm{d}x.}∫043π(sin6x+cos6x)2dx.
∫034πsin6xcos6x dx=112{\displaystyle \int_0^{\frac{3}{4} \pi} \sin 6 x \cos 6 x \, \mathrm{d}x = \frac{1}{12}}∫043πsin6xcos6xdx=121. ∫034π(sin6x+cos6x)2 dx=16+34π{\displaystyle \int_0^{\frac{3}{4} \pi} (\sin 6 x + \cos 6 x)^2 \, \mathrm{d}x = \frac{1}{6} + \frac{3}{4} \pi}∫043π(sin6x+cos6x)2dx=61+43π.