Question

1, \(\int\dfrac{dx}{1+\sqrt{x}}\)

2, \(\int\dfrac{sinx.cos^3x}{1+sin^2x}dx\)

Answer 1

\(I=\int\dfrac{dx}{1+\sqrt{x}}\)

Đặt \(\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2t.dt\)

\(\Rightarrow I=\int\dfrac{2t.dt}{1+t}=\int\left(2-\dfrac{2}{1+t}\right)dt=2t-2ln\left|1+t\right|+C\)

\(=2\sqrt{x}-2ln\left|1+\sqrt{x}\right|+C\)

2/

\(I=\int\dfrac{sinx.cos^3xdx}{1+sin^2x}=\int\dfrac{cos^3x.sinxdx}{2-cos^2x}\)

Đặt \(cosx=t\Rightarrow sinxdx=-dt\)

\(\Rightarrow I=\int\dfrac{t^3dt}{t^2-2}=\int\left(t+\dfrac{2t}{t^2-2}\right)dt=\int t.dt+\int\dfrac{2t.dt}{t^2-2}\)

\(=\int t.dt+\int\dfrac{d\left(t^2-2\right)}{t^2-2}=\dfrac{t^2}{2}+ln\left|t^2-2\right|+C\)

\(=\dfrac{cos^2x}{2}+ln\left|cos^2x-2\right|+C\)