Tích phân [Lần 6]

Bài 1. Tính tích phân $$\int_{0}^{\frac{\pi }{6}}\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$$

Đặt: $$J=\int_{0}^{\frac{\pi }{6}}\frac{\sin^3 x}{\sin^3 x+\cos^3 x}dx$$
Ta có:
$$I-J=\int_{0}^{\frac{\pi }{6}}\frac{\cos^3x-\sin^3 x}{\sin^3 x+\cos^3 x}dx=\int_{0}^{\frac{\pi }{6}}\frac{(\cos x-\sin  x)(1+\sin x.\cos x)}{\cos x+\sin  x)(1-\sin x.\cos x)}dx$$$$=-\int_{0}^{\frac{\pi }{6}}\frac{1+\sin x.\cos x}{(\cos x+\sin  x)(1-\sin x.\cos x)}d(\cos x+\sin x)$$
Đặt: $t=\cos x+\sin x\Rightarrow \sin x.\cos x=\dfrac{t^2-1}{2}$
Vậy:
$$I-J=-\int\limits_1^{\dfrac{1+\sqrt{3}}{2}} \dfrac{1+\dfrac{t^2-1}{2}}{t(1-\dfrac{t^2-1}{2})}dt=-\int\limits_1^{\dfrac{1+\sqrt{3}}{2}} \dfrac{1+t^2}{t(3-t^2)}dt$$$$=-\int\limits_1^{\dfrac{1+\sqrt{3}}{2}} \dfrac{t}{3-t^2}dt-\int\limits_1^{\dfrac{1+\sqrt{3}}{2}} \dfrac{t}{t^2(3-t^2)}dt$$$$=-\dfrac{1}{3}\left(\ln\bigg|t\bigg|-2\ln\bigg|3-t^2\bigg|\right)\bigg|_1^{\dfrac{1+\sqrt{3}}{2}} $$$$=\ln{\left(\dfrac{2.\sqrt[3]{43+27\sqrt{3}}}{13^{\frac{2}{3}}}\right)}$$
Mặt khác:
$$I+J=\int_{0}^{\frac{\pi }{6}}\frac{\sin^3 x+\cos ^3x}{\sin^3 x+\cos^3 x}dx=\int_{0}^{\frac{\pi }{6}}dx=x\bigg|_{0}^{\frac{\pi }{6}}=\dfrac{\pi}{6}$$.
Vậy:
$$I=\dfrac{1}{2}.\ln{\left(\dfrac{2.\sqrt[3]{43+27\sqrt{3}}}{13^{\frac{2}{3}}}\right)} + \dfrac{\pi}{12}$$

Bài 2. Tính tích phân: $$\displaystyle \int_{-1}^1 \frac{xe^x+e^x+1}{x^2(e^x+1)^2}\ dx$$

$$\displaystyle \int_{-1}^1 \frac{d(x.e^x+x)}{(xe^x+x)^2}\ dx =  - \frac{1}{x.e^x+x} \bigg|_{-1}^1$$

Bài 3. Tính tích phân: $$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1-\sin x}{(1+\cos x)e^x}\ dx$$

Viết lại: $$I=\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{e^{-x}}{1+\cos x}\ dx-\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\sin xe^{-x}}{1+\cos x}\ dx$$$$=\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{e^{-x}dx}{2\cos^2\dfrac{x}{2}}-\displaystyle \int_{0}^{\frac{\pi}{2}}\tan\dfrac{x}{2}. e^{-x}dx$$

Mặt khác ta có: $$\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{e^{-x}dx}{2\cos^2\dfrac{x}{2}}=\displaystyle \int_{0}^{\frac{\pi}{2}} e^{-x}d\left(\tan\dfrac{x}{2}\right)=e^{-x}\tan\dfrac{x}{2}\Bigg|_0^{\frac{\pi}{2}}+ \displaystyle \int_{0}^{\frac{\pi}{2}}\tan\dfrac{x}{2}. e^{-x}dx$$

Thay vào ta có: $I=e^{-x}\tan\dfrac{x}{2}\Bigg|_0^{\frac{\pi}{2}}=e^{-\frac{\pi}{2}}$

Bài 4. Tính tích phân: $$\int_{o}^{\pi }\frac{x\sin x}{1+\cos ^2 x}dx$$

Đặt $$x = \pi  - t \Rightarrow dx =  - dt$$
Ta được $$I =  - \int_\pi ^0 {\frac{{(\pi  - t)\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt = \pi \int_0^\pi  {\frac{{\sin x}}{{1 + c{\rm{o}}{{\rm{s}}^2}x}}}dx  - I$$
 $$\Rightarrow I = \frac{\pi }{2}\int_0^\pi  {\frac{{\sin x}}{{1 + c{\rm{o}}{{\rm{s}}^2}x}}} dx
$$

Bài 5. $$I=\displaystyle \int_1^4 \sqrt{\dfrac{1}{4x}+\dfrac{\sqrt{x}+e^x}{\sqrt{x}.e^{2x}}}dx$$

Ta có  : $\ \dfrac{1}{4x} + \dfrac{\sqrt x +e^x}{\sqrt x e^{2x}} =\dfrac{1}{4x} + \dfrac{1}{\sqrt x e^x} + \dfrac{1}{e^{2x}} = \left(\dfrac{1}{2\sqrt x} + \dfrac{1}{e^x} \right)^2$
Vậy ta có $\ \sqrt{\dfrac{1}{4x} + \dfrac{\sqrt x +e^x}{\sqrt x e^{2x}}}=\sqrt{\left(\dfrac{1}{2\sqrt x} + \dfrac{1}{e^x} \right)^2}$
Vậy ta có tích phân đã cho trở thành : $\ I =\displaystyle \int_{1}^{4} \left(\dfrac{1}{2\sqrt x} + \dfrac{1}{e^x} \right) \mbox{d}x = \left. \sqrt{x} - \dfrac{1}{e^x} \right|_{1}^{4} = 1 + \dfrac{1}{e} - \dfrac{1}{e^4}$