Integrals

$I={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\frac{xdx}{si{n}^4\left(2x\right)+co{s}^4\left(2x\right)}}$

           Let 2x = t then   $dx=\frac{1}{2}dt$

$I={\displaystyle \underset{}{\overset{}{\int }}\frac{\frac{t}{2}\cdot \frac{1}{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{tdt}{si{n}^{4}t+co{s}^{4}t}dt}$            

$I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-t\right)dt}{si{n}^{4}t+co{s}^{4}t}dt}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{si{n}^{4}tdt}{ta{n}^{4}t+1}}$            

Let tan t = y then

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+y^2\right)dy}{1+y^4}}$             

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2}dy}$

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+\frac{1}{y^2}\right)dy}{2+{\left(y-\frac{1}{y}\right)}^2}}$             

Let $y-\frac{1}{y}=u$  

$2I=\frac{\pi }{8}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2+u^2}}$

$=\frac{\pi }{8\sqrt[]{2}}{\left[ta{n}^{-1}\frac{4}{\sqrt[]{2}}\right]}_{-\infty }^{\infty }$                  

$I=\frac{{\pi }^2}{16\sqrt[]{2}}$

$k=\frac{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]-\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$            

k = 3

$\underset{x\to 0}{lim}{\left[si{n}^2\left(\frac{\pi }{2-3x}\right)\right]}^{se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$e^{\underset{x\to 0}{lim}\left[si{n}^2\left(\frac{\pi }{2-3x}\right)-1\right]se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$=e^{\underset{x\to 0}{lim}\frac{-si{n}^2\left(\frac{-3\pi x}{2\left(2-3x\right)}\right)}{si{n}^2\left(\frac{-5\pi x}{2\left(2-5x\right)}\right)}}=e^{-\frac{9}{25}}$

The integral is I = ∫ [ (x²-1) + tan? ¹ (x + 1/x)] / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
This is a complex integral. The provided solution splits it into two parts:
I? = ∫ (x²-1) / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
I? = ∫ 1 / (x? +3x²+1) dx
The solution proceeds with substitutions which are hard to follow due to OCR quality, but it seems to compare the final result with a given form to find coefficients α, β, γ, δ. The final expression shown is:
10 (α + βγ + δ) = 10 (1 + (1/2√5)*√5 + 1/2) seems incorrect.
The calculation is shown as 10 (1 + 1/10 - 1/2) = 10 (11/10 - 5/10) = 10 (6/10) = 6.

The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.

The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

I = ∫ (e²? + 2e? - e? - 1)e^ (e? +e? ) dx
I = ∫ (e²? + e? - 1)e^ (e? +e? ) dx + ∫ (e? - e? )e^ (e? +e? ) dx
I = ∫ (e? + 1 - e? )e^ (e? +e? ) dx + e^ (e? +e? )
(e? - e? + 1)dx = du
I = e^ (e? +e? ) + e^ (e? +e? ) = e^ (e? +e? ) (e? + 1) then g (x) = e? + 1
g (0) = 2

∫ (x/ (xsinx+cosx)²dx = ∫ (xcosx⋅xcosx)/ (xsinx+cosx)² dx
= x/cosx (-1/ (xsinx+cosx) + ∫ (cosx-xsinx)/cosx² (1/ (xsinx+cosx) dx
= -xsecx/ (xsinx+cosx) + ∫sec²xdx
= -xsecx/ (xsinx+cosx) + tanx + C

Below are a few important tips to remember the integrals of some particular functions:
1. Know the derivatives for each integral.
2. Make yourself familiar with the standard formulas first.
3. Practice daily for better memory.
4. Group similar formulas

$\mathrm{}4\alpha {\int }_{-1}^0?e^{\alpha x}dx+{\int }_0^2?e^{-\alpha x}dx=5$

$\Rightarrow 4\alpha \left(\frac{1-{\mathrm{e}}^{-\alpha }}{\alpha }-\frac{{\mathrm{e}}^{-2\alpha }-1}{-\alpha }\right)=5$

Let ${\mathrm{e}}^{-\alpha }=\mathrm{t}$

$\therefore 4{\mathrm{t}}^2+4\mathrm{t}-3=0$

$\Rightarrow \mathrm{t}=\frac{1}{2}$

$\alpha =\mathrm{l}\mathrm{n}?2$