Question

\(\int\left(\frac{1}{1+sinx}\right)dx\)

\(\int\left(sin^4x\right)dx\)

\(\int\left(sin^6x+cos^6x\right)dx\)

Answer 1

Lời giải:

Câu 1:

\(A=\int\frac{dx}{1+\sin x}=\int \frac{(1-\sin x)dx}{1-\sin^2 x}=\int\frac{(1-\sin x)dx}{\cos ^2x}=\int\frac{dx}{\cos ^2x}-\int\frac{\sin x dx}{\cos^2 x}\)

\(\Leftrightarrow A=\int d(\tan x)+\int\frac{d(\cos x)}{\cos^2 x}=\tan x-\frac{1}{\cos x}+c\)

Câu 2:

\(B=\int \sin ^4 xdx=\int \sin^2 x(1-\cos ^2x)dx=\int \sin^2 xdx-\int \sin^2 x\cos^2xdx\)

Ta thấy \(\int \sin^2xdx=\frac{1}{2}\int (1-\cos 2x)dx=\frac{x}{2}-\frac{\sin 2x}{4}+c\)

Và \(\int \sin ^2x\cos^2xdx=\frac{1}{4}\int \sin^22xdx=\frac{1}{8}\int (1-\cos4x)dx=\frac{x}{8}-\frac{\sin 4x}{32}+c\)

\(\Rightarrow B=\frac{3}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+c\)

Answer 2

Câu 3:

\(C=\int (\sin ^6 x+\cos^6 x)dx=\int (\sin^2x+\cos^2x)[\sin^4x-\sin^2x\cos^2x+\cos^4x)dx\)

\(\Leftrightarrow C=\int [(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x]dx\)

\(\Leftrightarrow C=\int dx-\frac{3}{4}\int\sin^22xdx=\int dx-\frac{3}{8}\int (1-\cos 4x)dx\)

\(\Leftrightarrow C=x-\frac{3x}{8}+\frac{3\sin 4x}{32}+c=\frac{5x}{8}+\frac{3\sin 4x}{32}+c\)