How do you find $\int_{\frac{π}{24}}^{\frac{π}{18}} 36 (1 + 3^{6 sec (6 x)}) (sec (6 x) tan (6 x)) d x$ $int_(pi/24)^(pi/18) 36(1+3^(6sec(6x)))(sec(6x)tan(6x))dx$ ?

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How do you find $\int_{\frac{π}{24}}^{\frac{π}{18}} 36 (1 + 3^{6 sec (6 x)}) (sec (6 x) tan (6 x)) d x$ $int_(pi/24)^(pi/18) 36(1+3^(6sec(6x)))(sec(6x)tan(6x))dx$ ?

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Answer 1 · 2017-01-17T16:22:54

$12 - 6 \sqrt{2} + \frac{3^{12} - 3^{6 \sqrt{2}}}{ln 3}$ $12-6sqrt2+(3^12-3^(6sqrt2))/ln3$

Explanation:

We will take this step-by-step. The first substitution we will use is:

$u = 6 x$ $u=6x$
$d u = 6 . d x$ $du=6color(white).dx$

The transformation of bounds becomes:

$x = \frac{π}{18} \Rightarrow u = 6 x = 6 (\frac{π}{18}) = \frac{π}{3}$ $x=pi/18" "=>" "u=6x=6(pi/18)=pi/3$

$x = \frac{π}{24} \Rightarrow u = 6 x = 6 (\frac{π}{24}) = \frac{π}{4}$ $x=pi/24" "=>" "u=6x=6(pi/24)=pi/4$

So:

$36 \int_{π / 24}^{π / 18} (1 + 3^{6 sec (6 x)}) (sec (6 x) tan (6 x)) d x$ $36int_(pi//24)^(pi//18)(1+3^(6sec(6x)))(sec(6x)tan(6x))dx$

$= 6 \int_{π / 24}^{π / 18} (1 + 3^{6 sec (6 x)}) (sec (6 x) tan (6 x)) (6 . d x)$ $=6int_(pi//24)^(pi//18)(1+3^(6sec(6x)))(sec(6x)tan(6x))(6color(white).dx)$

$= 6 \int_{π / 4}^{π / 3} (1 + 3^{6 sec (u)}) (sec (u) tan (u)) d u$ $=6int_(pi//4)^(pi//3)(1+3^(6sec(u)))(sec(u)tan(u))du$

Now, let:

$v = 6 sec (u)$ $v=6sec(u)$
$d v = 6 sec (u) tan (u) d u$ $dv=6sec(u)tan(u)du$

The bounds become:

$u = \frac{π}{3} \Rightarrow v = 6 sec (u) = \frac{6}{cos (\frac{π}{3})} = \frac{6}{\frac{1}{2}} = 12$ $u=pi/3" "=>" "v=6sec(u)=6/cos(pi/3)=6/(1/2)=12$

$u = \frac{π}{4} \Rightarrow v = sec (u) = \frac{6}{cos (\frac{π}{4})} = \frac{6}{\frac{1}{\sqrt{2}}} = 6 \sqrt{2}$ $u=pi/4" "=>" "v=sec(u)=6/cos(pi/4)=6/(1/sqrt2)=6sqrt2$

Then the integral becomes:

$= \int_{π / 4}^{π / 3} (1 + 3^{6 sec (u)}) (6 sec (u) tan (u) d u)$ $=int_(pi//4)^(pi//3)(1+3^(6sec(u)))(6sec(u)tan(u)du)$

$= \int_{6 \sqrt{2}}^{12} (1 + 3^{v}) d v$ $=int_(6sqrt2)^12(1+3^v)dv$

Splitting up the integral:

$= \int_{6 \sqrt{2}}^{12} d v + \int_{6 \sqrt{2}}^{12} 3^{v} d v$ $=int_(6sqrt2)^12dv+int_(6sqrt2)^12 3^vdv$

We can evaluate the first integral easily:

$= v ∣_{6 \sqrt{2}}^{12} + \int_{6 \sqrt{2}}^{12} 3^{v} d v$ $=v| _ (6sqrt2)^12 + int _(6sqrt2)^12 3^vdv$

$= 12 - 6 \sqrt{2} + \int_{6 \sqrt{2}}^{12} 3^{v} d v$ $=12-6sqrt2 + int _(6sqrt2)^12 3^vdv$

For the remaining integral, note that $3 = e^{ln 3}$ $3=e^ln3$ so $3^{v} = {(e^{ln 3})}^{v} = e^{v ln 3}$ $3^v=(e^ln3)^v=e^(vln3)$ .

$= 12 - 6 \sqrt{2} + \int_{6 \sqrt{2}}^{12} e^{v ln 3} d v$ $=12-6sqrt2+int_(6sqrt2)^12e^(vln3)dv$

Performing another substitution:

$w = v ln 3$ $w=vln3$
$d w = ln 3 . d v$ $dw=ln3color(white).dv$

The bounds:

$v = 12 \Rightarrow w = v ln 3 = 12 ln 3$ $v=12" "=>" "w=vln3=12ln3$

$v = 6 \sqrt{2} \Rightarrow w = v ln 3 = 6 \sqrt{2} (ln 3)$ $v=6sqrt2" "=>" "w=vln3=6sqrt2(ln3)$

Then:

$= 12 - 6 \sqrt{2} + \frac{1}{ln 3} \int_{6 \sqrt{2}}^{12} e^{v ln 3} (ln 3 . d v)$ $=12-6sqrt2+1/ln3int_(6sqrt2)^12e^(vln3)(ln3color(white).dv)$

$= 12 - 6 \sqrt{2} + \frac{1}{ln 3} \int_{6 \sqrt{2} (ln 3)}^{12 ln 3} e^{w} d w$ $=12-6sqrt2+1/ln3int_(6sqrt2(ln3))^(12ln3)e^wdw$

The integral of $e^{w}$ $e^w$ is itself:

$= 12 - 6 \sqrt{2} + \frac{1}{ln 3} (e^{w}) {∣ ∣ ∣}_{6 \sqrt{2} (ln 3)}^{12 ln 3}$ $=12-6sqrt2+1/ln3(e^w)|_(6sqrt2(ln3))^(12ln3)$

$= 12 - 6 \sqrt{2} + \frac{1}{ln 3} e^{12 ln 3} - \frac{1}{ln 3} e^{6 \sqrt{2} (ln 3)}$ $=12-6sqrt2+1/ln3e^(12ln3)-1/ln3e^(6sqrt2(ln3))$

$= 12 - 6 \sqrt{2} + \frac{1}{ln 3} {(e^{ln 3})}^{12} - \frac{1}{ln 3} {(e^{ln 3})}^{6 \sqrt{2}}$ $=12-6sqrt2+1/ln3(e^ln3)^12-1/ln3(e^ln3)^(6sqrt2)$

$= 12 - 6 \sqrt{2} + \frac{3^{12} - 3^{6 \sqrt{2}}}{ln 3}$ $=12-6sqrt2+(3^12-3^(6sqrt2))/ln3$

$\approx 473563.931508$ $approx473563.931508$

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How do you find $\int_{\frac{π}{24}}^{\frac{π}{18}} 36 (1 + 3^{6 sec (6 x)}) (sec (6 x) tan (6 x)) d x$ $int_(pi/24)^(pi/18) 36(1+3^(6sec(6x)))(sec(6x)tan(6x))dx$ ?

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