Integral of $$$- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}$$$ with respect to $$$x$$$

The calculator will find the integral/antiderivative of $$$- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}$$$ with respect to $$$x$$$, with steps shown.

Related calculator: Definite and Improper Integral Calculator

Please write without any differentials such as $$$dx$$$, $$$dy$$$ etc.
Leave empty for autodetection.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.

Your Input

Find $$$\int \left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)\, dx$$$.

The trigonometric functions expect the argument in radians. To enter the argument in degrees, multiply it by pi/180, e.g. write 45° as 45*pi/180, or use the appropriate function adding 'd', e.g. write sin(45°) as sind(45).

Solution

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- 9 i n t \sec{\left(2 \right)}$$$ and $$$f{\left(x \right)} = x \sin{\left(3 x \right)}$$$:

$${\color{red}{\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x}}} = {\color{red}{\left(- 9 i n t \sec{\left(2 \right)} \int{x \sin{\left(3 x \right)} d x}\right)}}$$

For the integral $$$\int{x \sin{\left(3 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=\sin{\left(3 x \right)} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\sin{\left(3 x \right)} d x}=- \frac{\cos{\left(3 x \right)}}{3}$$$ (steps can be seen »).

The integral can be rewritten as

$$- 9 i n t \sec{\left(2 \right)} {\color{red}{\int{x \sin{\left(3 x \right)} d x}}}=- 9 i n t \sec{\left(2 \right)} {\color{red}{\left(x \cdot \left(- \frac{\cos{\left(3 x \right)}}{3}\right)-\int{\left(- \frac{\cos{\left(3 x \right)}}{3}\right) \cdot 1 d x}\right)}}=- 9 i n t \sec{\left(2 \right)} {\color{red}{\left(- \frac{x \cos{\left(3 x \right)}}{3} - \int{\left(- \frac{\cos{\left(3 x \right)}}{3}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{3}$$$ and $$$f{\left(x \right)} = \cos{\left(3 x \right)}$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} - {\color{red}{\int{\left(- \frac{\cos{\left(3 x \right)}}{3}\right)d x}}}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} - {\color{red}{\left(- \frac{\int{\cos{\left(3 x \right)} d x}}{3}\right)}}\right)$$

Let $$$u=3 x$$$.

Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

The integral becomes

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\cos{\left(3 x \right)} d x}}}}{3}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{3}\right)$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{3}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}}{3}\right)$$

The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{9}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\sin{\left(u \right)}}}}{9}\right)$$

Recall that $$$u=3 x$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left({\color{red}{u}} \right)}}{9}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{9}\right)$$

Therefore,

$$\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x} = - 9 i n t \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left(3 x \right)}}{9}\right) \sec{\left(2 \right)}$$

Simplify:

$$\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x} = i n t \left(3 x \cos{\left(3 x \right)} - \sin{\left(3 x \right)}\right) \sec{\left(2 \right)}$$

Add the constant of integration:

$$\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x} = i n t \left(3 x \cos{\left(3 x \right)} - \sin{\left(3 x \right)}\right) \sec{\left(2 \right)}+C$$

Answer

$$$\int \left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)\, dx = i n t \left(3 x \cos{\left(3 x \right)} - \sin{\left(3 x \right)}\right) \sec{\left(2 \right)} + C$$$A


Please try a new game Rotatly