4. Под знак дифференциала постепенно загоняем: сначала косинус, затем двойку и наконец единицу, т.е.
![\frac{1}{2}d(2sinx+1) =cosxdx](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B2%7Dd%282sinx%2B1%29+%3Dcosxdx)
, чтобы получился табличный интеграл от степенной функции.
![\int\limits^{ \pi /2}_0 { \sqrt{2sinx+1} * cosx} \, dx =\int\limits^{ \pi /2}_0 { \sqrt{2sinx+1} } \, d(sinx) = \\ \\ =\int\limits^{ \pi /2}_0 { \frac{1}{2} \sqrt{2sinx+1} } \, d(2sinx+1) = \frac{1}{2}\int\limits^{ \pi /2}_0 {(2sinx+1)^{ \frac{1}{2} } } \, d(2sinx+1) = \\ \\ = \frac{1}{2} \frac{2}{3} (2sinx+1)^{ \frac{3}{2}}= \frac{1}{3} (2sinx+1)^{ \frac{3}{2}}|_{0}^{\pi /2}= \\ \\ = \frac{1}{3} (2sin \frac{ \pi }{2} +1)^{ \frac{3}{2}} -\frac{1}{3} (2sin 0 +1)^{ \frac{3}{2}} =](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Csqrt%7B2sinx%2B1%7D+%2A+cosx%7D+%5C%2C+dx+%3D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Csqrt%7B2sinx%2B1%7D+%7D+%5C%2C+d%28sinx%29+%3D+%5C%5C++%5C%5C+%3D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Cfrac%7B1%7D%7B2%7D+%5Csqrt%7B2sinx%2B1%7D+%7D+%5C%2C+d%282sinx%2B1%29+%3D+%5Cfrac%7B1%7D%7B2%7D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B%282sinx%2B1%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+%7D+%5C%2C+d%282sinx%2B1%29+%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B2%7D++%5Cfrac%7B2%7D%7B3%7D+%282sinx%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D%3D+%5Cfrac%7B1%7D%7B3%7D+%282sinx%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D%7C_%7B0%7D%5E%7B%5Cpi+%2F2%7D%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B3%7D+%282sin+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+-%5Cfrac%7B1%7D%7B3%7D+%282sin+0+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%3D)
![= \frac{1}{3} (2sin \frac{ \pi }{2} +1)^{ \frac{3}{2}} -\frac{1}{3} (2sin 0 +1)^{ \frac{3}{2}} = \frac{1}{3} \sqrt{27} -\frac{1}{3}= \sqrt{3} -\frac{1}{3}](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B1%7D%7B3%7D+%282sin+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+-%5Cfrac%7B1%7D%7B3%7D+%282sin+0+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%3D+%5Cfrac%7B1%7D%7B3%7D+%5Csqrt%7B27%7D+-%5Cfrac%7B1%7D%7B3%7D%3D+%5Csqrt%7B3%7D++-%5Cfrac%7B1%7D%7B3%7D)
5.
![\int\limits^4_2 { \frac{1}{x-1} } \, dx =\int\limits^4_2 { \frac{1}{x-1} } \, d(x-1) =ln(x-1)|_{2}^{4}=ln3-ln1=ln3](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E4_2+%7B+%5Cfrac%7B1%7D%7Bx-1%7D+%7D+%5C%2C+dx+%3D%5Cint%5Climits%5E4_2+%7B+%5Cfrac%7B1%7D%7Bx-1%7D+%7D+%5C%2C+d%28x-1%29+%3Dln%28x-1%29%7C_%7B2%7D%5E%7B4%7D%3Dln3-ln1%3Dln3)
<span>Log 0,5 (0,2х+6)>-3
ОДЗ
</span>0,2х+6>0
x>-30
![log_{0.5} (0.2x+6)\ \textgreater \ log_{0.5} (0.5)^{-3} \\ \\ 0.2x+6\ \textgreater \ (0.5)^{-3} \\ \\ 0.2x+6\ \textgreater \ 8 \\ \\ x\ \textgreater \ 10](https://tex.z-dn.net/?f=log_%7B0.5%7D+%280.2x%2B6%29%5C+%5Ctextgreater+%5C+log_%7B0.5%7D+%280.5%29%5E%7B-3%7D+%5C%5C++%5C%5C++0.2x%2B6%5C+%5Ctextgreater+%5C+%280.5%29%5E%7B-3%7D+%5C%5C++%5C%5C+0.2x%2B6%5C+%5Ctextgreater+%5C+8+%5C%5C++%5C%5C+x%5C+%5Ctextgreater+%5C+10)
Решением будет система из логарифма и ОДЗ. Но надо посмотреть на основание. Если а>1 то знак не меняем, если а<1 то знак меняем.
![\left \{ {{x\ \textgreater \ -30} \atop {x\ \textless \ \ 10}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+-30%7D+%5Catop+%7Bx%5C+%5Ctextless+%5C++%5C+10%7D%7D+%5Cright.+)
x∈(-30;10)
<span>2√х-20=0
√х=20:2
√х=10
х=10²=100</span>
4.75 / 100%= 0.0475
0.0475 * 75% = 3.56 м