How do you integrate #int (2sinx+3cosx)dx#?

Question

Scarlett Allison · Accepted Answer

#3sinx-2cosx+C# #int (2sinx+3cosx)dx= 2*int sinx dx + 3* int cosx dx# Applying standard integrals: #= 2*(-cosx) +3*sinx +C# #=3sinx-2cosx+C#

Scarlett Allison · Answer

undefined To integrate $ \int (2\sin(x) + 3\cos(x)) \, dx $, you can use integration by parts. Let $ u = 2\sin(x) $ and $ dv = dx $, then differentiate $ u $ to find $ du $, and integrate $ dv $ to find $ v $. Then use the formula $ \int u \, dv = uv - \int v \, du $.

$ u = 2\sin(x) $  
$ du = 2\cos(x) \, dx $  
$ dv = dx $  
$ v = x $

Using integration by parts formula:

$ \int (2\sin(x) + 3\cos(x)) \, dx = 2\sin(x)x - \int x(2\cos(x)) \, dx $

Now, we need to integrate $ \int x(2\cos(x)) \, dx $. This can be done by integration by parts again.

Let $ u = x $ and $ dv = 2\cos(x) \, dx $.

$ du = dx $  
$ v = 2\sin(x) $

Using integration by parts formula again:

$ \int x(2\cos(x)) \, dx = x(2\sin(x)) - \int 2\sin(x) \, dx $

Now integrate $ \int 2\sin(x) \, dx $ directly.

$ \int 2\sin(x) \, dx = -2\cos(x) $

Plugging this back into the original integral:

$ \int (2\sin(x) + 3\cos(x)) \, dx = 2\sin(x)x - \left(x(2\sin(x)) - \int 2\sin(x) \, dx \right) $

$ = 2\sin(x)x - \left(x(2\sin(x)) - (-2\cos(x)) \right) $

$ = 2\sin(x)x - 2x\sin(x) + 2\cos(x) + C $

where $ C $ is the constant of integration.