Consider the following integral.
Find a substitution to rewrite the integrand as
.
Evaluate the given integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.)
Problem:
Evaluate the integral:∫cot(3x) dx\int \cot(3x)\,dx∫cot(3x)dx
Correct Answer:
∫cot(3x) dx=13ln∣sin(3x)∣+C\int \cot(3x)\,dx = \frac{1}{3} \ln\left|\sin(3x)\right| + C∫cot(3x)dx=31ln∣sin(3x)∣+C
Explanation
To evaluate the integral ∫cot(3x) dx\int \cot(3x)\,dx∫cot(3x)dx, we begin by recalling the identity:cot(u)=cos(u)sin(u)\cot(u) = \frac{\cos(u)}{\sin(u)}cot(u)=sin(u)cos(u)
So the integral becomes:∫cot(3x) dx=∫cos(3x)sin(3x) dx\int \cot(3x)\,dx = \int \frac{\cos(3x)}{\sin(3x)}\,dx∫cot(3x)dx=∫sin(3x)cos(3x)dx
This form suggests using substitution. Let:u=sin(3x)u = \sin(3x)u=sin(3x)
Then, compute the derivative of uuu with respect to xxx:dudx=3cos(3x)⇒du=3cos(3x) dx\frac{du}{dx} = 3\cos(3x) \Rightarrow du = 3\cos(3x)\,dxdxdu=3cos(3x)⇒du=3cos(3x)dx
Solving for dxdxdx:cos(3x) dx=13 du\cos(3x)\,dx = \frac{1}{3}\,ducos(3x)dx=31du
Now, return to the integral and substitute:∫cos(3x)sin(3x) dx=∫1u⋅13 du=13∫1u du\int \frac{\cos(3x)}{\sin(3x)}\,dx = \int \frac{1}{u} \cdot \frac{1}{3}\,du = \frac{1}{3} \int \frac{1}{u}\,du∫sin(3x)cos(3x)dx=∫u1⋅31du=31∫u1du
This is a standard integral:13ln∣u∣+C\frac{1}{3} \ln|u| + C31ln∣u∣+C
Now substitute back for u=sin(3x)u = \sin(3x)u=sin(3x):13ln∣sin(3x)∣+C\frac{1}{3} \ln|\sin(3x)| + C31ln∣sin(3x)∣+C
Thus, the final answer is:13ln∣sin(3x)∣+C\boxed{\frac{1}{3} \ln\left|\sin(3x)\right| + C}31ln∣sin(3x)∣+C
This result is correct and reflects both the chain rule and substitution method. The provided screenshot shows incorrect substitution steps for u=sin(x)u = \sin(x)u=sin(x) and du=1x dxdu = \frac{1}{x} \, dxdu=x1dx, which don’t match the given integrand. However, the final answer box is correct, so the evaluation was done properly despite the substitution error in the earlier step.
