Question

Bằng cách biến đổi các hàm số lượng giác, hãy tính :

a) \(\int\sin^4xdx\)

b) \(\int\dfrac{1}{\sin^3x}dx\)

c) \(\int\sin^3x\cos^4xdx\)

d) \(\int\sin^4x\cos^4xdx\)

e) \(\int\dfrac{1}{\cos x\sin^2x}dx\)

g) \(\int\dfrac{1+\sin x}{1+\cos x}dx\)

Answer 1

a) \(\sin^4x=\left(\sin^2x\right)^2=\left(\dfrac{1-\cos2x}{2}\right)^2\)

\(=\dfrac{1}{4}\left(1-2\cos2x+\cos^22x\right)\)

\(=\dfrac{1}{4}\left(1-2.\cos2x+\dfrac{1+\cos4x}{2}\right)\)

\(=\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\)

Vậy:

\(\int\sin^4x\text{dx}=\int\left(\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\right)\text{dx}\)

\(=\dfrac{3}{8}x-\dfrac{1}{4}\sin2x+\dfrac{1}{32}\sin4x+C\)