Tính tích phân dx/4(sinx)^2+9(cosx)^2

Đáp án:

$\int\dfrac{dx}{4sin^2x+9cos^2x}$$=\dfrac{1}{6}.arctan \dfrac{2tanx}{3}+C$

Giải thích các bước giải:

$\int\dfrac{dx}{4sin^2x+9cos^2x}$

$=\int\dfrac{1}{(4\dfrac{sin^2x}{cos^2x}+9)cos^2x}dx$

$=\int\dfrac{1}{(4\dfrac{sin^2x}{cos^2x}+9)}.\dfrac{1}{cos^2x}dx$

$=\int\dfrac{1}{(4tan^2x+9)}.\dfrac{1}{cos^2x}dx$

$=\int\dfrac{1}{(4tan^2x+9)}d(tanx)$

$=\int\dfrac{1}{2}.\dfrac{1}{((2tanx)^2+9)}d(2tanx)$

$=\dfrac{1}{2}.\dfrac{1}{3}.arctan \dfrac{2tanx}{3}+C$

$=\dfrac{1}{6}.arctan \dfrac{2tanx}{3}+C$

Áp dụng $\int \dfrac{dx}{x^2+a^2}=\dfrac{1}{a}.arctan\dfrac{x}{a}+C$