# Can you please solve these questions involving integrals? Thanks.

## $\int 9 \cos \left(5 x - 3\right) \mathrm{dx}$ $\int 4 \sin \left(\frac{x}{3}\right) \mathrm{dx}$ $\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx}$ $\int \left(\sec 3 x \tan 3 x - 15 \cos 3 x\right) \mathrm{dx}$ $\int \left(4 \csc 10 x \cot 10 x + 2 {\sec}^{2} 10 x\right) \mathrm{dx}$ I'm asking for your help in answering these questions for my advanced Math class. I'm okay with integrals in general, but I tend to get confused when it comes to trigonometric functions. Thank you!

Feb 20, 2018

$1.$

$\int 9 \cos \left(5 x - 3\right) \mathrm{dx} = \frac{9}{5} \int \cos \left(5 x - 3\right) d \left(5 x - 3\right) = \frac{9}{5} \sin \left(5 x - 3\right) + C$

$2.$

$\int 4 \sin \left(\frac{x}{3}\right) \mathrm{dx} = 12 \int \sin \left(\frac{x}{3}\right) d \left(\frac{x}{3}\right) = - 12 \cos \left(\frac{x}{3}\right) + C$

$3.$

$\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx} = \int \left(\frac{1}{{\cos}^{2} 7 x} - \frac{1}{{\sin}^{2} 7 x}\right) \mathrm{dx}$

$\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx} = \int \left(\frac{{\sin}^{2} 7 x - {\cos}^{2} 7 x}{{\cos}^{2} 7 x {\sin}^{2} 7 x}\right) \mathrm{dx}$

$\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx} = - 4 \int \frac{\cos 14 x}{{\sin}^{2} 14 x} \mathrm{dx}$

$\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx} = - \frac{2}{7} \int \frac{d \left(\sin 14 x\right)}{{\sin}^{2} 14 x}$

$\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx} = \frac{2}{7 \sin 14 x} + C$

or alternatively:

$\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx} = \frac{1}{7} \int {\sec}^{2} \left(7 x\right) d \left(7 x\right) - \frac{1}{7} \int {\csc}^{2} \left(7 x\right) d \left(7 x\right)$

$\int \left({\sec}^{2} 7 x - {\csc}^{2} 7 x\right) \mathrm{dx} = \frac{\tan \left(7 x\right) + \cot \left(7 x\right)}{7} + C$

$4.$

$\int \left(\sec 3 x \tan 3 x - 15 \cos 3 x\right) \mathrm{dx} = \frac{1}{3} \int \sec 3 x \tan 3 x d \left(3 x\right) - 5 \int \cos \left(3 x\right) d \left(3 x\right)$

$\int \left(\sec 3 x \tan 3 x - 15 \cos 3 x\right) \mathrm{dx} = \sec \frac{3 x}{3} - 5 \sin \left(3 x\right)$

$5.$

$\int \left(4 \csc \left(10 x\right) \cot \left(10 x\right) + 2 {\sec}^{2} \left(10 x\right)\right) \mathrm{dx} = \frac{2}{5} \int \csc \left(10 x\right) \cot \left(10 x\right) d \left(10 x\right) + \frac{1}{5} \int {\sec}^{2} \left(10 x\right) d \left(10 x\right)$

$\int \left(4 \csc \left(10 x\right) \cot \left(10 x\right) + 2 {\sec}^{2} \left(10 x\right)\right) \mathrm{dx} = \frac{- 2 \csc \left(10 x\right) + \tan \left(10 x\right)}{5}$

Feb 20, 2018

1. $\frac{9}{5} \sin \left(5 x - 3\right) + C$ 2. $- 12 \cos \left(\frac{x}{3}\right)$+C 3. $\frac{1}{7} \tan \left(7 x\right) + \frac{1}{7} \cot \left(7 x\right) + C$ 4. $\frac{1}{3} \sec \left(3 x\right) - 5 \sin \left(3 x\right) + C$ 5. $- \frac{2}{5} \csc \left(10 x\right) - \frac{1}{5} \tan \left(10 x\right) + C$

#### Explanation:

1. u=5x-3
du=5dx
(1/5)du=dx

Rewrite:
$\frac{9}{5} \int \cos \left(u\right) \mathrm{du}$ (Bring all constants outside of the integral).
$\frac{9}{5} \sin \left(u\right) + C$
$\frac{9}{5} \sin \left(5 x - 3\right) + C$
2. u=x/3
du=dx/3
3du=dx

Rewrite:
$\left(4\right) \left(3\right) \int \sin \left(u\right) \mathrm{du}$
$12 \int \sin \left(u\right) \mathrm{du}$
$- 12 \cos \left(u\right) + C$
$- 12 \cos \left(\frac{x}{3}\right)$+C
3. Split up the integral:
$\int {\sec}^{2} \left(7 x\right) \mathrm{dx} - \int {\csc}^{2} \left(7 x\right) \mathrm{dx}$

Make the following substitution for both integrals:

u=7x
du=7dx
(1/7)du=dx

$\frac{1}{7} \int {\sec}^{2} \left(u\right) \mathrm{du} - \frac{1}{7} \int {\csc}^{2} \left(u\right) \mathrm{du}$

Now this is only composed of basic integrals:
$\frac{1}{7} \tan \left(u\right) + \frac{1}{7} \cot \left(u\right) + C$
$\frac{1}{7} \tan \left(7 x\right) + \frac{1}{7} \cot \left(7 x\right) + C$
4. Split up the integral:
$\int \sec \left(3 x\right) \tan \left(3 x\right) \mathrm{dx} - \int 15 \cos \left(3 x\right) \mathrm{dx}$

Make the following substitution for both integrals:
u=3x
du=3dx
(1/3)du=dx

$\frac{1}{3} \int \sec \left(u\right) \tan \left(u\right) \mathrm{du} - \frac{15}{3} \int \cos \left(u\right) \mathrm{du}$

$\frac{1}{3} \int \sec \left(u\right) \tan \left(u\right) \mathrm{du} - 5 \int \cos \left(u\right) \mathrm{du}$

Now we have an integral composed of only basic integrals:

$\frac{1}{3} \sec \left(u\right) - 5 \sin \left(u\right) + C$

$\frac{1}{3} \sec \left(3 x\right) - 5 \sin \left(3 x\right) + C$
5. Split up the integral:
$\int 4 \csc \left(10 x\right) \cot \left(10 x\right) \mathrm{dx} - \int 2 {\sec}^{2} \left(10 x\right) \mathrm{dx}$

Make the following substitution for both integrals:

u=10x
du=10dx
(1/10)du=dx
$\frac{4}{10} \int \csc \left(u\right) \cot \left(u\right) \mathrm{du} - \frac{2}{10} \int {\sec}^{2} \left(u\right) \mathrm{du}$

$\frac{2}{5} \int \csc \left(u\right) \cot \left(u\right) \mathrm{du} - \frac{1}{5} \int {\sec}^{2} \left(u\right) \mathrm{du}$

Now this is only composed of basic integrals:

$- \frac{2}{5} \csc \left(u\right) - \frac{1}{5} \tan \left(u\right) + C$

$- \frac{2}{5} \csc \left(10 x\right) - \frac{1}{5} \tan \left(10 x\right) + C$