Tính tích phân \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\sqrt{1+3\cos x}}dx\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\sqrt{1+3\cos x}}dx=-\frac{3}{2}\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\sqrt{1+3\cos x}}dx=\frac{3}{2}\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\sqrt{1+3\cos x}}dx=\frac{2}{3}\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\sqrt{1+3\cos x}}dx=-\frac{2}{3}\) Hướng dẫn giải: Đặt \(t=\sqrt{1+3\cos x}\) \(\Rightarrow1+3\cos x=t^2\) => \(-3\sin x\text{d}x=2t\text{d}t\) Đổi cận: \(x|^{\frac{\pi}{2}}_0\Rightarrow t|^1_2\) \(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\sqrt{1+3\cos x}}dx=\int\limits^1_2\frac{-\frac{2}{3}t\text{d}t}{t}=\frac{2}{3}\int\limits^2_1\text{d}t\) \(=\frac{2}{3}t|^2_1=\frac{4}{3}-\frac{2}{3}=\frac{2}{3}\)
Tính tích phân \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}dx\) \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}dx=\frac{4+3\sqrt{2}}{4}\) \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}dx=\frac{-4+3\sqrt{2}}{4}\) \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}dx=\frac{4-3\sqrt{2}}{4}\) \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}dx=\frac{-4-3\sqrt{2}}{4}\) Hướng dẫn giải: Ta có: \(\sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(\sin x-\cos x\right)\) \(\sin2x+2\left(1+\sin x+\cos x\right)=2\sin x\cos x+2+\sin x+\cos x\) \(=\sin^2x+\cos^2x+1+2\sin x\cos x+2\sin x+2\cos x=\left(1+\sin x+\cos x\right)^2\) \(I=\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}dx=\int\limits^{\frac{\pi}{4}}_0\frac{\frac{\sqrt{2}}{2}\left(\sin x-\cos x\right)}{\left(1+\sin x+\cos x\right)^2}\text{d}x\) Ta đặt \(t=1+\sin x+\cos x\) => \(\text{d}t=\left(\cos x-\sin x\right)\text{d}x=-\left(\sin x-\cos x\right)\text{d}x\) \(I=\int\limits^{1+\sqrt{2}}_2\frac{\frac{\sqrt{2}}{2}\left(-\text{d}t\right)}{t^2}\) \(=\frac{\sqrt{2}}{2}\frac{1}{t}|^{1+\sqrt{2}}_2\) \(=\frac{\sqrt{2}}{2}\left[\frac{1}{1+\sqrt{2}}-\frac{1}{2}\right]\) \(=\frac{4-3\sqrt{2}}{4}\)
Tính \(\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}dx\) ? \(\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}dx=-\frac{5\sqrt{3}}{9}+\frac{1}{2}ln\left(2+\sqrt{3}\right)\) \(\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}dx=-\frac{10\sqrt{3}}{27}+\frac{1}{2}ln\left(2+\sqrt{3}\right)\) \(\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}dx=\frac{10\sqrt{3}}{9}+\frac{1}{2}ln\left(2+\sqrt{3}\right)\) \(\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}dx=-\frac{10\sqrt{3}}{9}+\frac{1}{2}ln\left(2+\sqrt{3}\right)\) Hướng dẫn giải: Ta có: \(\cos2x=\cos^2x-\sin^2x=\cos^2x\left(1-\tan^2x\right)\) \(I=\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}dx=\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\left(1-\tan^2x\right)}.\frac{1}{\cos^2x}\text{d}x\) Đặt \(t=\tan x\Rightarrow\text{d}t=\frac{1}{\cos^2x}\text{d}x\) \(I=\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{t^4}{1-t^2}\text{d}t\) \(=\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{t^4-1+1}{1-t^2}\text{d}t\) \(=\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{\left(t^2-1\right)\left(t^2+1\right)+1}{1-t^2}\text{d}t\) \(=-\int\limits^{\frac{1}{\sqrt{3}}}_0\left(t^2+1\right)\text{d}t+\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{1}{\left(1-t\right)\left(1+t\right)}\text{d}t\) \(=-\left(\frac{t^3}{3}+t\right)|^{\frac{1}{\sqrt{3}}}_0+\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{1}{2}\left[\frac{1}{1-t}+\frac{1}{1+t}\right]\text{d}t\) \(=-\left(\frac{t^3}{3}+t\right)|^{\frac{1}{\sqrt{3}}}_0+\frac{1}{2}\left[\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{-\text{d}\left(1-t\right)}{1-t}+\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{\text{d}\left(1+t\right)}{1+t}\right]\) \(=-\left(\frac{t^3}{3}+t\right)|^{\frac{1}{\sqrt{3}}}_0+\frac{1}{2}\left[-\ln\left(1-t\right)+\ln\left(1+t\right)\right]|^{\frac{1}{\sqrt{3}}}_0\) \(=\left[-\left(\frac{t^3}{3}+t\right)+\frac{1}{2}\ln\frac{1+t}{1-t}\right]|^{\frac{1}{\sqrt{3}}}_0\) \(=-\frac{10\sqrt{3}}{27}+\frac{1}{2}ln\left(2+\sqrt{3}\right)\)
Tính tích phân \(\int\limits^4_0\frac{4x-1}{\sqrt{2x+1}+1}dx\) \(\int\limits^4_0\frac{4x-1}{\sqrt{2x+1}+1}dx=\frac{10}{3}+\ln2\) \(\int\limits^4_0\frac{4x-1}{\sqrt{2x+1}+1}dx=\frac{22}{3}+\ln2\) \(\int\limits^4_0\frac{4x-1}{\sqrt{2x+1}+1}dx=\frac{-22}{3}+\ln2\) \(\int\limits^4_0\dfrac{4x-1}{\sqrt{2x+1}+1}dx=\dfrac{-10}{3}+\ln2\) Hướng dẫn giải: Đặt \(t=\sqrt{2x+1}+1\) => \(\left(t-1\right)^2=2x+1\) => \(2\left(t-1\right)\text{d}t=2\text{d}x\) và \(x=\frac{\left(t-1\right)^2-1}{2}\) \(I=\int\limits^4_0\frac{4x-1}{\sqrt{2x+1}+1}dx=\int\limits^4_2\frac{4\frac{\left(t-1\right)^2-1}{2}}{t}\left(t-1\right)\text{d}t\) \(=\int\limits^4_2\frac{\left(2t^2-4t-1\right)\left(t-1\right)}{t}\text{d}t\) \(=\int\limits^4_2\frac{2t^3-6t^2+3t+1}{t}\text{d}t\) \(=\int\limits^4_2\left(2t^2-6t+3+\frac{1}{t}\right)\text{d}t\) \(=\left(\frac{2t^3}{3}-\frac{6t^2}{2}+3t+\ln t\right)|^4_2\) \(=\frac{22}{3}+\ln2\)
Tính tích phân \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx=\frac{2}{5}\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx=\frac{27}{23}\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx=\frac{34}{27}\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx=\frac{35}{29}\) Hướng dẫn giải: \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx=\int\limits^{\frac{\pi}{2}}_0\frac{2\sin x.\cos x+\sin x}{\sqrt{1+3\cos x}}\text{d}x=\int\limits^{\frac{\pi}{2}}_0\frac{\left(2\cos x+1\right)\sin x\text{d}x}{\sqrt{1+3\cos x}}\) Đặt \(t=\sqrt{1+3\cos x}\) => \(t^2=1+3\cos x\) => \(2t\text{d}t=-3\sin x\text{d}x\) và \(\cos x=\frac{t^2-1}{3}\) Thay vào ta được tích phân đa thức của t: \(\int\limits^1_2\frac{\left(2.\frac{t^2-1}{3}+1\right).\left(-\frac{2}{3}\right)t\text{d}t}{t}\) \(=\int\limits^2_1\frac{2}{9}\left(2t^2+1\right)\text{d}t\) \(=\frac{2}{9}\left(2.\frac{t^3}{3}+t\right)|^2_1\) \(=\frac{34}{29}\)
Dùng phép đổi biến số \(\dfrac{\sqrt{1+x^2}}{x}=t\) hãy chuyển tích phân \(I=\int\limits^{\sqrt{3}}_1\frac{\sqrt{1+x^2}dx}{x^2}\) về tích phân biến t. \(I=\int\limits^{\frac{2}{\sqrt{3}}}_{\sqrt{2}}\frac{tdt}{t^2-1}\) \(I=\int\limits^3_2\frac{tdt}{t^2+1}\) \(I=-\int\limits^{\frac{2}{\sqrt{3}}}_{\sqrt{2}}\frac{tdt}{t^2-1}\) \(I=\int\limits^3_2\frac{tdt}{t^2+1}\) Hướng dẫn giải: Đặt \(\dfrac{\sqrt{1+x^2}}{x}=t\Rightarrow\dfrac{1+x^2}{x^2}=t^2\Rightarrow\dfrac{1}{x^2}=t^2-1\) \(\Rightarrow x^2=\frac{1}{t^2-1}\Rightarrow2xdx=\frac{-2tdt}{\left(t^2-1\right)^2}\) . Do đó \(\frac{\sqrt{1+x^2}}{x^2}dx=t.\frac{xdx}{x^2}=-t.\frac{\frac{tdt}{\left(t^2-1\right)^2}}{\frac{1}{t^2-1}}=-\frac{t^2dt}{t^2-1}\) và \(I=-\int\limits^{\frac{2}{\sqrt{3}}}_{\sqrt{2}}\frac{tdt}{t^2-1}\) .
