How do you evaluate the integral #int tan(3x-5)dx#?

Answer 1

# 1/3ln|sec(3x-5)|+C.#

If we subst. #(3x-5)=t," then, "3dx=dt.#
#:. I=inttan(3x-5)dx=1/3inttan(3x-5)(3)dx,#
#=1/3inttan t dt,#
#=1/3ln|sec t|,#
# rArr I=1/3ln|sec(3x-5)|+C.#
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Answer 2

#\frac{ln|sec(3x - 5)|}{3} + C#

We're going to use u-substitution:

#u = 3x - 5# #du = 3 dx#

Keep in mind the above statement also means that:

#\frac{1}{3}du = dx#

After u-substitution the integral becomes:

#\int \frac{1}{3}tan(u) du#
We can take out the #\frac{1}{3}#:
#\frac{1}{3} \int tan(u) du#

Which we know is equal to:

#\frac{1}{3}ln|sec(u)| + C#

Now you just plug in the original value of u:

#\frac{1}{3}ln|sec(3x - 5)| + C#

and that's your answer :P

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Answer 3

The answer is #=-1/3ln|cos(3x-5)|+C#

We perform this integral by substitution

Let #u=3x-5#, #=>#, #du=3dx#
#inttan(3x-5)dx=1/3inttanudu#
#=1/3int(sinudu)/cosu#
Let #v=cosu#, #=>#, #dv=-sinudu#
#1/3int(sinudu)/cosu=-1/3int(dv)/v#
#=-1/3ln|v|#
#=-1/3ln|cosu|#
#=-1/3ln|cos(3x-5)|+C#
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Answer 4

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Substituting these into the integral, we get ∫tan(u) * (1/3)To evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cosTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) duTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(uTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du.To evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)|To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. NowTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| +To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now,To evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + CTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, weTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C.To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrateTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back inTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) withTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in forTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect toTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for uTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to uTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u:To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u,To evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, whichTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equalsTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cosTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -lnTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|To evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cosTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(uTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - 5)|To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(u)| +To evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - 5)|/3To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(u)| + CTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - 5)|/3 +To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(u)| + C,To evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - 5)|/3 + CTo evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(u)| + C, where CTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - 5)|/3 + C.To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(u)| + C, where C isTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - 5)|/3 + C.To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(u)| + C, where C is theTo evaluate the integral of tan(3x - 5) dx, you can use the substitution method. Let u = 3x - 5, then du = 3 dx. Rewrite the integral in terms of u: int tan(u) * (1/3) du. The integral of tan(u) du is -ln|cos(u)| + C. Substitute back in for u: -ln|cos(3x - 5)|/3 + C.To evaluate the integral ∫tan(3x - 5) dx, we use the substitution method. Let u = 3x - 5. Then, du/dx = 3 and dx = du/3. Substituting these into the integral, we get ∫tan(u) * (1/3) du. Now, we integrate tan(u) with respect to u, which equals -ln|cos(u)| + C, where C is the constant of integration. Finally, substituting back u = 3x - 5, we have -ln|cos(3x - 5)|/3 + C as the result.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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