Calculate the following integral:
∫ sec^4 (3x)/ tan^3 (3x) dx
For this one, can I bring up the tan to tan^-3?
yes, so you have
∫sec^4(3x)* tan^-3(3x) dx
∫sec^2(3x)*(tan^2(3x)+1)*tan^-3(3x) dx
Now let u = tan(3x)
du = 3sec^2(3x) dx
∫1/u + 1/u^3 du/3
= 1/3 ln(u) - 1/6 u^-2
= 1/3 ln(tan(3x) - 1/6 cot^2(3x) + C
thanks Steve! You're a lifesaver
Yes, you can rewrite the integral using negative exponents.
Let's rewrite tan^3 (3x) as (tan (3x))^3 = tan^(-3) (3x).
Now, the integral becomes:
∫ sec^4 (3x) / tan^(-3) (3x) dx
To solve the integral ∫sec^4(3x) / tan^3(3x) dx, you can rewrite the expression using trigonometric identities. One way to proceed is by manipulating the trigonometric functions to express everything in terms of either secant or tangent.
We know that sec^2(x) = 1 + tan^2(x). From this identity, we can rewrite sec^4(x) as:
sec^4(x) = (1 + tan^2(x))^2.
Now let's rewrite the integral using this identity:
∫ sec^4(3x) / tan^3(3x) dx = ∫ (1 + tan^2(3x))^2 / tan^3(3x) dx.
To simplify the integral further, we can use substitution. Let's substitute u = tan(3x) and calculate du:
du = 3sec^2(3x) dx,
dx = du / (3sec^2(3x)).
Now we can rewrite the integral using the substitution:
∫ (1 + tan^2(3x))^2 / tan^3(3x) dx = ∫ (1 + u^2)^2 / (u^3) (du / (3sec^2(3x))).
Simplifying the integral, we get:
∫ (1 + u^2)^2 / (3u^3) sec^2(3x) du.
Now we have the integral in terms of u and du. To proceed further, you can simplify the expression at this point by expanding (1 + u^2)^2 and applying the power rule for integration with rational exponents.