Question

Evaluate the integral using integration by parts with the indicated choices of u and dv. (Use C for the constant of integration.) 9θ cos(θ) dθ ; u = 9θ, dv = cos(θ) dθ

Answer 1

Answer:

$\int9\theta cos\theta d\theta= 9\theta sin\theta +9 cos\theta +C$

Step-by-step explanation:

We are given that a function

$f(\theta)=9\thetacos\theta$

Where u=$9\theta$ and v=$cos\theta$

We have to find the integral using by parts integration formula

$\int u\cdot vdt = u\int v dt-\int(\frac{du}{dt}\int vdt )dt$

$\int 9\theta cos\theta d\theta=9\theta\int cos\theta d\theta -\int (\frac{dcos\theta}{d\theta}\cdot\int cos\theta d\theta$

$\int 9\thetacos\theta \d\theta=9\theta sin\theta-9(-cos\theta)+C$

Where C is integration constant.

$\int 9\theta cos\theta d\theta =9\theta \cdot sin\theta -9\int sin\theta d\theta +C$

Using $\int cos\thetad\theta=sin\theta$

$\int9\theta cos\theta d\theta= 9\theta sin\theta +9 cos\theta +C$

Using $\int sin\thetad\theta =-cos\theta$.

Therefore, the integration of the given function

$\int9\theta cos\theta d\theta= 9\theta sin\theta +9 cos\theta +C$