不定积分求教 ∫ 1/(sinx^2 + 2cosx^2 ) dx
人气:360 ℃ 时间:2020-06-02 09:28:24
解答
∫ 1 / (sin²x + 2 cos²x) dx
=∫ (1 / cos²x) / (2 + tan²x) dx
令u = tanx,du = sec²x dx
=∫ 1 / (2 + u²) du
令u = √2 tanv,du = √2 sec²v dv
2 + u² = 2 + 2 tan²v = 2 sec²v
=(√2 / 2) ∫ 1 / sec²v * sec²v dv
=1 / √2 * v+C
=(1 / √2) arctan(u / √2) + C
=(1 / √2) arctan[tanx / √2] + C
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