Maths Integrals

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

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

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

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

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

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

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

  ${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$           

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$        

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$        

I = ∫ (cos 4x-1)/ (cot x-tan x) dx = ∫ (2 cos² 2x-1-1)/ (cos²x-sin²x)/ (sin x cos x) dx
= 2 ∫ (cos² 2x-1) sin x cos x) / (cos 2x) dx = ∫ (cos² 2x-1) sin 2x) / (cos 2x) dx
Put cos 2x = t
-2 sin 2x dx = dt
I = -1/2 ∫ (t²-1)/t dt = 1/2 ∫ (1/t - t) dt
= 1/2 (ln|t| - t²/2) + c
= 1/2 ln|cos 2x| - 1/4 cos² (2x) + c

I = ∫? ² [log? x] dx
By using graph
I = ∫? 0 dx + ∫? ² 1 dx
I = 0 + e² - e

∫? ^ (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

sin? ¹ (√3/2) + cos? ¹ (-√3/2) + tan? ¹ (-1)
= π/3 + 5π/6 – π/4
= (4π+10π–3π)/12 = 11π/12

(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

I = ∫? ² (x³+|x|)/ (e|x|+1) dx . (i)
I = ∫? ² (x³+|x|)/ (e? |x|+1) dx . (ii)
= ∫? ² |x| dx = 2∫? ² x dx
= [x²/2]? ² = (16/4 + 4/2) - 0
= 4+2=6