Chapter 17: Problem 6

Find $\int 3 \tan 2 x+2 \sin 3 x \mathrm{~d} x$.

Short Answer

Expert verified

Answer: The antiderivative of the function $3 \tan 2x + 2 \sin 3x$ is $-\frac{1}{2} \ln|\cos(2x)| - \frac{1}{3} \cos(3x) + C$, where C is a constant of integration.

Step by step solution

Rewrite the Function

We start by rewriting the function in terms of simplified trigonometric functions. $\tan(x) = \frac{\sin(x)}{\cos(x)}$, so we can rewrite the function as: $$\int (3 \cdot \frac{\sin(2x)}{\cos(2x)} + 2 \sin (3x)) \mathrm{~d} x$$

Find Antiderivative of $\sin(2x)/\cos(2x)$

To find the antiderivative of $\frac{\sin(2x)}{\cos(2x)}$, let $u = 2x$ and $\mathrm{d}u = 2 \mathrm{~d}x$. Then we have: $$ \int \frac{\sin(u)}{\cos(u)} \frac{1}{2} \mathrm{d}u$$ Now, we can use the substitution $v = \cos(u)$ and $\mathrm{d}v = -\sin(u) \mathrm{~d}u$. This gives us: $$-\frac{1}{2} \int \frac{1}{v} \mathrm{d}v$$ Now, we can integrate: $$ -\frac{1}{2} \ln|v| + C = -\frac{1}{2} \ln|\cos(u)| + C = -\frac{1}{2} \ln|\cos(2x)| + C $$

Find Antiderivative of $\sin(3x)$

To find the antiderivative of $\sin(3x)$, we use the simple substitution $w = 3x$ and $\mathrm{d}w = 3 \mathrm{~d}x$. Then we have: $$ \int \sin(w) \frac{1}{3} \mathrm{d}w$$ Now we can integrate: $$ -\frac{1}{3} \cos(w) + C = -\frac{1}{3} \cos(3x) + C$$

Combine the Antiderivatives

Now we can combine the antiderivatives from Steps 2 and 3 to find the antiderivative of the original function: $$ \int 3 \tan 2x + 2 \sin 3x \, \mathrm{d}x = -\frac{1}{2} \ln|\cos(2x)| - \frac{1}{3} \cos(3x) + C$$ where C is a constant of integration.

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Find \(\int 3 \tan 2 x+2 \sin 3 x \mathrm{~d} x\).

Short Answer

Step by step solution

Rewrite the Function

Find Antiderivative of \(\sin(2x)/\cos(2x)\)

Find Antiderivative of \(\sin(3x)\)

Combine the Antiderivatives

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