Question

Помогите пожалуйста решить задачу ​

Answer 1

Ответ:

$1)\displaystyle \int\limits_{-\pi/2}^{\pi /2}\, sin^2\frac{x}{4}\cdot cos^4 \frac{x}{4}\, dx=\int\limits_{-\pi/2}^{\pi /2}\, (sin\frac{x}{4}\cdot cos\frac{x}{4})^2\cdot cos^2\frac{x}{4}\, dx=\\\\\\=\int\limits_{-\pi/2}^{\pi /2}\, (\frac{1}{2}\cdot sin\frac{x}{2})^2\cdot cos^2\frac{x}{4}\, dx=\frac{1}{4}\int\limits^{\pi/2}_{-\pi /2}\, sin^2\frac{x}{2}\cdot cos^2\frac{x}{4}\, dx=$

$\displaystyle =\frac{1}{4}\int\limits^{-\pi/2}_{-\pi /2}\, \frac{1-cosx}{2}\cdot \frac{1+cos\frac{x}{2}}{2}\, dx=\frac{1}{16}\int\limits^{\pi/2}_{-\pi /2}\, (1-cosx)\cdot (1+cos\frac{x}{2})\, dx=\\\\\\=\frac{1}{16}\int\limits^{\pi/2}_{-\pi /2}\, (1+cos\frac{x}{2}-cosx-cosx\cdot cos\frac{x}{2})\, dx=\\\\\\=\frac{1}{16}\cdot (x+2sin\frac{x}{2}-sinx)\Big|_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\frac{1}{16}\int\limits^{\pi/2}_{-\pi /2}\, \frac{1}{2}\cdot (cos\frac{3x}{2}+cos\frac{x}{2})\, dx=$

$\displaystyle =\frac{1}{16}\cdot (\ \frac{\pi}{2}+\frac{\pi}{2}+2sin\frac{\pi}{4}+2sin\frac{\pi}{4}-sin\frac{\pi}{2}-sin\frac{\pi }{2}}\ )-\\\\\\-\frac{1}{32}\cdot (\ \frac{2}{3}sin\frac{3x}{2}+2sin\frac{x}{2}\ )\Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{1}{16}\cdot (\pi +2\sqrt2-2\ )-\frac{1}{32}\cdot (\ \frac{2}{3}+\frac{2}{3}+2+2\ )=\\\\\\=\frac{\pi}{16}+\frac{\sqrt2-1}{8}-\frac{1}{6}=\frac{\pi }{16}+\frac{3\sqrt2-7}{24}$

$2)\ \ \displaystyle \int \, \frac{\sqrt{(1+x^2)^3}}{x^2}\, dx=\Big[\ x=tgt\ ,\ dx=\frac{dt}{cos^2t}\ \Big]=\int \frac{\sqrt{(1+tg^2t)^3}}{tg^2t}\cdot \frac{dt}{cos^2t}=\\\\\\=\int \frac{\sqrt{(1/cos^2t)^3}}{sin^2t}\, dt=\int \frac{dt}{cos^3t\cdot sin^2t}=\int \frac{\overbrace{(sin^2t+cos^2t)^2}^{1}}{cos^3t\cdot sin^2t}\, dt=\\\\\\=\int \frac{sin^4t+2sin^2t\cdot cos^2t+cos^4t}{cos^3t\cdot sin^2t}\, dt=$

$\displaystyle =\int \frac{sin^4t}{cos^3t\cdot sin^2t}\, dt+\int \frac{2sin^2t\cdot cos^2t}{cos^3t\cdot sin^2t}\, dt+\int \frac{cos^4t}{cos^3t\cdot sin^2t}\, dt=\\\\\\=\int \frac{sin^2t}{cos^3t}\, dt+2\int \frac{dt}{cost}\, dt+\int \frac{cost}{sin^2t}\, dt=(*)$

$\displaystyle \star \int \frac{dt}{cost}=\int \frac{dt}{sin(\dfrac{\pi}{2}+t)}=\int \frac{dt}{2sin(\dfrac{\pi}{4}+\dfrac{t}{2})\cdot cos(\dfrac{\pi}{4}+\dfrac{t}{2})}=\\\\\\=\int \frac{dt}{2\cdot \dfrac{sin(\dfrac{\pi}{4}+\dfrac{t}{2})}{cos(\dfrac{\pi}{4}+\dfrac{t}{2})}\cdot cos^2(\dfrac{\pi}{4}+\dfrac{t}{2})}=\int \dfrac{dt}{2\cdot tg(\dfrac{\pi}{4}+\dfrac{t}{2})\cdot cos^2(\dfrac{\pi}{4}+\dfrac{t}{2})}=$

$\displaystyle =\int \frac{d( tg(\dfrac{\pi}{4}+\dfrac{t}{2}))}{ tg(\dfrac{\pi}{4}+\dfrac{t}{2})}=ln\Big|\, tg(\dfrac{\pi}{4}+\dfrac{t}{2})\, \Big|+C\ ;$

$\displaystyle \star \int \frac{sin^2t}{cos^3t}\, dt=\int \frac{sint\cdot sint\, dt}{cos^3t}=\Big[\ u=sint\ ,\ du=cost\, dt\ ,\ dv=\frac{sint\, dt}{cos^3t}\ ,\\\\\\v=\frac{1}{2cos^2t}\ \Big]=uv-\int v\, du=\frac{sint}{2cos^2t}-\int \frac{dt}{2cost}=\frac{sint}{2cos^2t}-\frac{1}{2}\, ln\Big|tg(\frac{t}{2}+\frac{\pi}{4})\Big|+C;$

$\displaystyle \star \ \int \frac{cost}{sin^2t}\, dt=\int \frac{d(sint)}{sin^2t}=-\frac{1}{sint}+C\ ;$  

$\displaystyle (*)=\frac{sint}{2cos^2t}+\frac{3}{2}\cdot ln\Big|tg(\frac{x}{2}+\frac{\pi}{4})\Big|-\frac{1}{sint}+C\ ;\\\\\\\int \limits _1^{\sqrt8}\frac{\sqrt{(1+x^2)^3}}{x^2}\, dx=\left(\frac{sin(arctgx)}{2cos^2(arctgx)}+\frac{3}{2}\cdot ln\Big|tg(\frac{arctgx}{2}+\frac{\pi}{4})\Big|-\frac{1}{sin(arctgx)}\right)\Big|_1^{\sqrt8}$

P.S. Этот интеграл можно ещё вычислить с помощью интегрирования биномиальных дифференциалов , 3 случай .