Question

Evaluate the interval (Calculus 2)

Answer 1

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\ \displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x \end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating \sec^2 kx }\\\\ \displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C}) \end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating \sec kx \tan kx }\\\\ \displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C}) \end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

https://brainly.com/question/27805589

https://brainly.com/question/28155016

Answer 2

Substitute $y=6x$ and $dy=6\,dx$ to transform the integral to

$\displaystyle \int \frac{12}{1-\sin(6x)} \, dx = 2 \int \frac{dy}{1 - \sin(y)}$

Now substitute $t=\tan\left(\frac y2\right)$ and $dt=\frac12 \sec^2\left(\frac y2\right) \, dy$ to transform this to

$\displaystyle 2 \int \frac{dy}{1 - \sin(y)} = 2 \int \frac1{1-\frac{2t}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} = 4 \int \frac{dt}{(t-1)^2}$

Finally, substitute $s = t-1$ and $ds=dt$ to get

$\displaystyle 4 \int \frac{dt}{(t-1)^2} = 4 \int \frac{ds}{s^2} = -\dfrac4s + C$

Now recover the antiderivative in terms of $x$.

$\displaystyle \int \frac{12}{1-\sin(6x)} \, dx = -\frac4s + C \\\\ ~~~~~~~~ = -\frac4{t-1} + C \\\\ ~~~~~~~~ = -\frac4{\tan\left(\frac y2\right) - 1} + C \\\\ ~~~~~~~~ = \boxed{-\frac4{\tan(3x) - 1} + C}$