Question

一道不定积分题，谢谢大神来解答

Answer 1

解：原式=∫[sec²x/(9+4tan²x)dx (分子分母乘sec²x)
=∫sec²xdx/(9+4tan²x)
=(1/6)∫(2/3)sec²xdx/(1+((2/3)tanx)²)
=(1/6)∫d((2/3)tanx)/(1+((2/3)tanx)²)
=(1/6)arctan((2/3)tanx)+C (C是积分常数)。

Answer 2

∫1/( 4 (sinx)^2 + 9 (cosx)^2 ) dx
=∫1/(cosx)^2 * 1/( 4 (tanx)^2 + 9 ) dx
=∫ 1/( 4 (tanx)^2 + 9 ) d(tanx)
=1/9∫ 1/( (2/3 tanx )^2 + 1 ) d(tanx)
=1/6 ∫ 1/( 1+ (2/3 tanx )^2 ) d(2/3 tanx)
=1/6 arctan(2/3 tanx ) +C