Maths Integrals

Kindly consider the following figure

I = ∫ from -π/2 to π/2 (1 / (1+e^ (sin x) dx
I = ∫ from -π/2 to π/2 (e^ (sin x) / (1+e^ (sin x) dx
2I = ∫ from -π/2 to π/2 1dx ⇒ I = 1/2 ∫ from -π/2 to π/2 dx
I = 1/2 [x] from -π/2 to π/2 ⇒ I = π/2

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

Clearly, ∫ [0 to n] {x}dx = n/2
∫ [0 to n] [x]dx = 1 + 2 + 3 . . . n − 1
= n (n-1)/2
∴ (n (n-1)/2)² = n/2 {10n (n-1)}
Solving, n = 21

∫ [π/6 to π/3] (d/dx (tan? x) * sin³3x + tan? x * d/dx (sin³3x) dx
= 1/2 ∫ [π/6 to π/3] d/dx (tan? x * sin?3x) dx
= 1/2 [tan? x · sin?3x] from π/6 to π/3
= 1/2 [ (√3)? · 0 - 1/ (√3)? )]
= -1/18

 f (x) = ∫ (from 1 to 3) (√x dx)/ (1+x)² = ∫ (from 1 to √3) (t⋅2tdt)/ (1+t²)² (put √x = t)
= [ (-t/ (1+t²)] (from 1 to √3) + [tan? ¹t] (from 1 to √3) [Applying by parts]
= (-√3/4 + 1/2) + (π/3 - π/4)
= (-√3+2)/4 + π/12

∫ (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