Use the method that follows to find the solution to the indefinite integral ∫ sin (3x+8) dx.
Find inside the integral the composition of two functions and let the u= inner function. In this example the inner function is (3x+8). So set (3x+8) = u.
Take the derivative of u= (3x+8) and you find that du= 3 dx. Then rearrange the derivative to get it in the form of du/3= dx.
Substitute into the integral in Step 1 everything that depends on x in terms of u. This gives ∫ sin (u) du/3 or ∫1/3 sin (u) du. Now this integral is easier to integrate.
Use the Table of Indefinite Integrals and see that ∫c f (u) du= c ∫ f (u) du, c is any constant and ∫sin (u) du= -cos (u) + C. In our case c=1/3.
Integrate ∫1/3 sin (u) du with respect to u using the information in Step 5 and you get that 1/3 ∫ sin (u) du = 1/3[-cos (u)] + C.
Substitute back in terms of the original variable x and since in Step 2 we set u= (3x+8) you get the final answer of 1/3[-cos (3x+8)] +C where C is a constant.