Question

Integration of sin(10x+3)dx.

Answer 1

Answer:

Let u=10x+3:

$\int \:sin\left(10x+3\right)dx=\int \:sin\left(u\right)\cdot \frac{1}{10}\cdot du=\frac{1}{10}\int \:sin\left(u\right)du=\frac{1}{10}\left(-cos\left(u\right)\right)+C=-\frac{1}{10}cos\left(10x+3\right)+C$

The key here is to use u-substitution, which is the integration version of chain rule.

Hope this helps :)

Step-by-step explanation: