Tính tích phân: $I=$$\int\limits {\frac{dx}{9sin^2x+4cos^2x}} \, dx$
2 câu trả lời
Lời giải:
Đặt $t=tgx=>dt=(1+tg^2x)dx$ hay $dx=\frac{dt}{1+t^2}$.Ta có:
$I=$$\int\limits {\frac{dx}{9sin^2x+4cos^2x}} \, dx=$ $\int\limits {\frac{\frac{dt}{1+t^2}}{\frac{9t^2}{1+t^2}+\frac{4}{1+t^2}}} \, dx=$ $\int\limits {\frac{dt}{((3t)^2+2^2}} \, =\frac{1}{3}$ $\int\limits {\frac{3dt}{(3t)^2+2^2}} \, =\frac{1}{3}.\frac{1}{2}.arctg\frac{3t}{2}+C=\frac{1}{6}.arctg(\frac{3}{2}tgx)+C$
Giải thích:
Đặt $t=tgx=>dt=(1+tg^2x)dx$ hay $dx=\frac{dt}{1+t^2}$.Ta có:
$I=$$\int\limits {\frac{dx}{9sin^2x+4cos^2x}} \, dx=$ $\int\limits {\frac{\frac{dt}{1+t^2}}{\frac{9t^2}{1+t^2}+\frac{4}{1+t^2}}} \, dx=$ $\int\limits {\frac{dt}{((3t)^2+2^2}} \, =\frac{1}{3}$ $\int\limits {\frac{3dt}{(3t)^2+2^2}} \, =\frac{1}{3}.\frac{1}{2}.arctg\frac{3t}{2}+C=\frac{1}{6}.arctg(\frac{3}{2}tgx)+C$
Chúc bạn học tốt!!!!