How do you find the indefinite integral of #int (4sinx)/(3tanx)dx#?

Question

Luke Alvarez · Accepted Answer

The answer is #=4/3sinx+C# We use #tanx=sinx/cosx# #(4sinx)/(3tanx)# #=(4sinx)/(3sinx/cosx)# #=4/3cosx# So, #int(4tanxdx)/(3cosx)=4/3intcosxdx# #=4/3sinx+C#

Christopher Allers · Answer

undefined To find the indefinite integral of $ \int \frac{4\sin(x)}{3\tan(x)} \, dx $, we can use the trigonometric identity $ \tan(x) = \frac{\sin(x)}{\cos(x)} $ to rewrite $ \frac{4\sin(x)}{3\tan(x)} $ as $ \frac{4\sin(x)}{3\left(\frac{\sin(x)}{\cos(x)}\right)} $. Then, we simplify by multiplying the numerator and denominator by $ \cos(x) $ to get $ \frac{4\sin(x)\cos(x)}{3\sin(x)} $. Now, $ \sin(x) $ cancels out, leaving $ \frac{4\cos(x)}{3} $. Thus, the indefinite integral becomes $ \int \frac{4\cos(x)}{3} \, dx $, which equals $ \frac{4}{3} \int \cos(x) \, dx $. Integrating $ \cos(x) $ yields $ \sin(x) + C $, where $ C $ is the constant of integration. Therefore, the indefinite integral of $ \int \frac{4\sin(x)}{3\tan(x)} \, dx $ is $ \frac{4}{3}\sin(x) + C $.

Emilia Aguilar · Answer

undefined To find the indefinite integral of $ \int \frac{4 \sin(x)}{3 \tan(x)} \, dx $, we can rewrite the expression in terms of sine and cosine functions:

$$ \tan(x) = \frac{\sin(x)}{\cos(x)} $$

Substituting $ \tan(x) $ with $ \frac{\sin(x)}{\cos(x)} $, we get:

$$ \frac{4 \sin(x)}{3 \tan(x)} = \frac{4 \sin(x)}{3 \left(\frac{\sin(x)}{\cos(x)}\right)} = \frac{4 \sin(x) \cos(x)}{3 \sin(x)} = \frac{4 \cos(x)}{3} $$

Now, the integral becomes:

$$ \int \frac{4 \cos(x)}{3} \, dx $$

This is a straightforward integral:

$$ \int \frac{4 \cos(x)}{3} \, dx = \frac{4}{3} \int \cos(x) \, dx = \frac{4}{3} \sin(x) + C $$

Where $ C $ is the constant of integration.

Therefore, the indefinite integral of $ \int \frac{4 \sin(x)}{3 \tan(x)} \, dx $ is $ \frac{4}{3} \sin(x) + C $.