Definite Integrals

$\text{27.(32)}\begin{array}{rr}& I=2{\int }_0^{\pi /2}??\left({\mathrm{c}\mathrm{o}\mathrm{s}}^3?x+{\mathrm{c}\mathrm{o}\mathrm{s}}^4?x+{\mathrm{s}\mathrm{i}\mathrm{n}}^4?x\right)dx\\ & I=2\left[{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^3?xdx+{\int }_0^{\pi /2}??{\mathrm{c}\mathrm{o}\mathrm{s}}^4?xdx+{\int }_0^{\pi /2}??{\mathrm{s}\mathrm{i}\mathrm{n}}^4?xdx\right]\\ & I=2\left[\frac{2}{3}\times 1+\frac{3\times 1}{4\times 2}\times \frac{\pi }{2}+\frac{3\times 1}{4\times 2}\times \frac{\pi }{2}\right]\\ & I=2\left[\frac{2}{3}+\frac{3\pi }{8}\right]\Rightarrow I=\frac{4}{3}+\frac{6\pi }{8}\Rightarrow 24I=32+18\pi \\ & 24I-18\pi =32\end{array}$

In the integral J, substitute x + 1 = t

$\Rightarrow dx=dtand{x}^2+2x=\left(t^2-1\right)$            

Now   $J={\displaystyle \underset{1}{\overset{e}{\int }}\frac{e^{\frac{t^2-2}{2}}}{t}}dtandK={\displaystyle \underset{1}{\overset{e}{\int }}t}lnt{e}^{\frac{t^2-2}{2}}dt$

Hence   $\left(J+K\right)={\displaystyle \underset{1}{\overset{e}{\int }}{e}^{\frac{t^2-2}{2}}}\left(\frac{1}{t}+tlnt\right)dt$

$={\left(e^{\frac{t^2-2}{2}}lnt\right)}_{t=1}^{t=e}=e^{\frac{e^2-2}{2}}={\left(\sqrt[]{e}\right)}^{e^2-2}$

I? = ∫? ¹ (1 − x? )? · 1 dx
= (1 - x? )? x|? ¹ - ∫? ¹ 7 (1 - x? )? (-4x³)xdx = -28∫? ¹ (1 − x? )? (1 − x? − 1)dx
I? = −28∫? ¹ (1 − x? )? dx + 28∫? ¹ (1 − x? )? dx
I? = -28I? + 28I?
29I? = 28I?
I? /I? = 28/29 ⇒ (29/4) * (I? /I? ) = (29/4) * (28/29) = 7

? C? +? C? =? C?
Σ (from r=0 to 3)? C? =? C? +? C? +? C? +? C?
=? C? +? C? +? C? +? C?
= ¹²C? = (12×11×10)/6 = 220

∫ (from 0 to π/2) (x/2)|cos (2x)|dx
∫ (from 0 to π/4) (x/2)cos (2x)dx - ∫ (from π/4 to π/2) (x/2)cos (2x)dx
= [x sin (2x)/4 - ∫sin (2x)/4 dx] (from 0 to π/4) - [x sin (2x)/4 - ∫sin (2x)/4 dx] (from π/4 to π/2)
= [x sin (2x)/4 + cos (2x)/8] (from 0 to π/4) - [x sin (2x)/4 + cos (2x)/8] (from π/4 to π/2)

${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$  

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ ->value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$        

$l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{e^{cosx}sinxdx}{\left(1+co{s}^{2}x\right)\left(e^{cosx}+e^{-cosx}\right)}}$

$2l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{sindx}{1+co{s}^{2}x}=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{sinxdx}{1+co{s}^{2}x}}}$

$l={\displaystyle \underset{1}{\overset{0}{\int }}\frac{-dt}{1+t^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}=\frac{\pi }{4}}}$

∫? ^ (π/2) sin³x e? sin²? dx
=∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
=2∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
Let cos²x=t⇒sin2xdx=−dt
=−2∫? (1−t)e? (¹? ) dt/ (−2cosx)
=1/e ∫? e? dt −∫? te? dt
=1/e [e? ]? − [te? −e? ]?
=2e−3∫? ¹ √t e? dt
⇒α=2, β=3
α+β=5

(cos (2π/7) + cos (4π/7) + cos (6π/7)/ (sin (3π/7)cos (π/7) = (2sin (π/7)cos (2π/7) + 2sin (π/7)cos (4π/7) + 2sin (π/7)cos (6π/7)/ (2sin (π/7)
= (sin (3π/7)-sin (π/7) + sin (5π/7)-sin (3π/7) + sin (π)-sin (5π/7)/ (2sin (π/7) = (-sin (π/7)/ (2sin (π/7) = -1/2