Tích phân [Lần 5]

Bài 1. Tính tích phân $$I=\int_0^{\dfrac{\pi }{2}} {\ln \dfrac{{{{\left( {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}^{1 + \cos x}}}}{{1 + \cos x}}} dx$$

Đặt: $x=\dfrac{\pi}{2}-t$
Thì:
$$I=\int_0^{\dfrac{\pi }{2}} {\ln \dfrac{{{{\left( {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}^{1 + \cos x}}}}{{1 + \cos x}}} dx=-\int^0_{\dfrac{\pi }{2}} \ln {\dfrac{\left(1+\cos t\right)^{1+\sin t}}{1+\sin t}}dt=\int_0^{\dfrac{\pi }{2}} \ln {\dfrac{\left(1+\cos t\right)^{1+\sin t}}{1+\sin t}}dt$$
Vậy:
$$2I=\int_0^{\dfrac{\pi }{2}} {\ln \dfrac{{{{\left( {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}^{1 + \cos x}}}}{{1 + \cos x}}} dx+\int_0^{\dfrac{\pi }{2}} \ln {\dfrac{\left(1+\cos x\right)^{1+\sin x}}{1+\sin x}}dx$$ $$=\int_0^{\dfrac{\pi }{2}} \ln (1+\sin x).(1+\cos x-1)+\ln(1+\cos x)(1+\sin x-1)dx$$ $$=\int_0^{\dfrac{\pi }{2}} \ln (1+\sin x).\cos x.dx+\int_0^{\dfrac{\pi }{2}} \ln (1+\cos x).\sin x.dx$$ $$=\int_0^{\dfrac{\pi }{2}} \ln (1+\sin x).d(1+\sin x)-\int_0^{\dfrac{\pi }{2}} \ln (1+\cos x).d(1+\cos x)$$ $$=\ln (1+\sin x).(1+\sin x)-(1+\sin x)-\ln (1+\cos x).(1+\cos x)+(1+\cos x) \bigg|^\dfrac{\pi}{2}_0$$

Bài 2. Tính tích phân sau: $$I=\int \sqrt[3]{\left( \sin (x)^3  +  \cos (x)^3  \right)\left( \sin (x) - \cos (x) \right)^3} dx$$

Ta có:
$$I=\int \sqrt[3]{\left( \sin^3x  +  \cos^3x \right)\left( \sin (x) - \cos (x) \right)^3} dx$$ $$=-\int \sqrt[3]{\left( \sin^3x  +  \cos^3x \right)}.d\left( \sin  +\cos x \right)$$ $$=-\int \sqrt[3]{\left( \sin x  +  \cos x \right).\left(1-\sin x.\cos x\right)}.d\left( \sin x +\cos x \right)$$
Đặt: $t=\left( \sin x +\cos x \right) \Rightarrow \sin x.\cos x=\dfrac{t^2-1}{2}$
Do đó:
$$I=-\int\sqrt[3]{t.\left(1-\dfrac{t^2-1}{2}\right)}dt$$ $$=\dfrac{1}{\sqrt[3]{2}}\int\sqrt[3]{t.\left(t^2-3\right)}dt$$

Bài 3. Tính tích phân: $$I=\displaystyle \int_1^2 \dfrac{x-1}{x^2\sqrt{2x^2-2x+1}}dx$$

Ta có:
$$I=\displaystyle \int_1^2 \dfrac{x-1}{x^2\sqrt{2x^2-2x+1}}dx$$ $$=\displaystyle \int_1^2 \dfrac{2x^2- x - (2x^2-2x+1)}{x^2\sqrt{2x^2-2x+1}}dx$$ $$=\displaystyle \int_1^2\dfrac{ \dfrac{x(2x-1)}{\sqrt{2x^2-2x+1}}dx- \sqrt{2x^2-2x+1}dx}{x^2}$$ $$=\displaystyle \int_1^2 \dfrac{xd(\sqrt{2x^2-2x+1})- \sqrt{2x^2-2x+1}dx}{x^2}$$ $$=\displaystyle \int_1^2 d\left(\dfrac{\sqrt{2x^2-2x+1}}{x}\right)(1)$$ $$=\left(\dfrac{\sqrt{2x^2-2x+1}}{x}\right)\bigg|_1^2$$ $$=\dfrac{(\sqrt{5}-2)}{2}.$$

Bài 4. Tính tích phân $$\int\limits_0^{\ln 3} {\frac{{{e^x} - 1}}{{{e^{2x}} + 1}}dx}$$

Ta có:
$$\int\limits_0^{\ln 3} \frac{e^x - 1}{e^{2x} + 1}dx=\int\limits_0^{\ln 3} \frac{e^{2x}-(e^{2x}+1)+e^x}{e^{2x} + 1}dx=\int\limits_0^{\ln 3} \frac{e^{2x}}{e^{2x} + 1}dx-\int\limits_0^{\ln 3} dx+\int\limits_0^{\ln 3} \frac{e^x}{e^{2x} + 1}dx$$
Mà:
$$\int\limits_0^{\ln 3} \frac{e^{2x}}{e^{2x} + 1}dx=\dfrac{1}{2}\int\limits_0^{\ln 3} \frac{1}{e^{2x} + 1}d(e^{2x})=\dfrac{1}{2}.\ln\bigg|e^{2x} + 1\bigg|\bigg|_0^{\ln 3}$$ $$\int\limits_0^{\ln 3} dx=x\bigg|_0^{\ln 3}$$ $$\int\limits_0^{\ln 3} \frac{e^x}{e^{2x} + 1}dx=\int\limits_0^{\ln 3} \frac{1}{e^{2x} + 1}d(e^x)$$
Đặt: $e^x=\tan t$
Ta sẽ thu được kết quả.

Bài 5. Tính: $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \sqrt{\tan x}dx$

Đặt: $t=\sqrt{\tan x} \Rightarrow t^2=\tan x \Rightarrow 2tdt=\dfrac{1}{\cos^2x}dx=(1+\tan^2x)dx=(1+t^4)dx$
Vậy:
$$I=\int \limits^{\sqrt{\sqrt{3}}}_1\dfrac{t.2t}{1+t^4}dt.=2.\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{t^2}{1+t^4}dt=\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{2}{\dfrac{1}{t^2}+t^2}dt.$$ $$=\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{1-\dfrac{1}{t^2}}{\left(\dfrac{1}{t}+t\right)^2-2}dt+\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{\dfrac{1}{t^2}+1}{\left(\dfrac{1}{t}-t\right)^2+2}dt$$
Mà:
$$\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{1-\dfrac{1}{t^2}}{\left(\dfrac{1}{t}+t\right)^2-2}dt=\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{1}{\left(\dfrac{1}{t}+t\right)^2-2}d\left(t+\dfrac{1}{t}\right)$$
$$\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{\dfrac{1}{t^2}+1}{\left(\dfrac{1}{t}-t\right)^2+2}dt=\int \limits^{\sqrt{\sqrt{3}}}_1 \dfrac{1}{\left(t-\dfrac{1}{t}\right)^2+2}d\left(t-\dfrac{1}{t}\right)$$