The integral ∫1+32sin2x(1−31)(cosx−sinx)dx is equal to
Solutions · 1
Official solution
Writing cosx−sinx=2sin(4π−x) and 1+32sin2x=32(sin3π+sin2x), then using sum-to-product on the denominator, the integrand reduces to a combination of csc terms:
\[\frac12\int\left[\csc\left(\frac\pi6+x\right)-\sec\left(\frac\pi6-x\right)\right]dx\]
Integrating each standard csc,sec form and combining logs gives
\[\frac12\log_e\left|\frac{\tan\left(\frac x2+\frac{\pi}{12}\right)}{\tan\left(\frac x2+\frac\pi6\right)}\right|+C\]