I'm doing trigonometric integrals
i wanted to know im doing step
is my answer right?
∫ tan^3 (2x) sec^5(2x) dx
=∫ tan^2(2x) sec^4(2x) tan*sec(2x) dx
=∫ (sec^2(2x)-1)sec^4 tan*sec(2x) dx
let u=sec x, du= 1/2 tan*sec(2x) dx
=1/2∫ (u^2(2x)-1) u^4 du
=1/2∫ (u^8(2x)-u^4) du
=1/2 sec^9/9-sec^5/5 +c
you dropped some 2's here and there, and the final integral is
1/2∫ (u^2-1) u^4 du
= 1/2 ∫ u^6 - u^4 du
= 1/2 (1/7 u^7 - 1/5 u^5)
= 1/14 sec^7(2x) - 1/10 sec^5(2x) + C
To verify if your answer for the integral ∫ tan^3 (2x) sec^5(2x) dx is correct, let's go through the steps together:
Step 1: Start with the original integral ∫ tan^3 (2x) sec^5(2x) dx.
Step 2: Rewrite tan^3 (2x) as tan^2 (2x) * tan(2x).
Step 3: Rewrite sec^5 (2x) as sec^4 (2x) * sec(2x).
Step 4: Now we have ∫ tan^2 (2x) sec^4 (2x) tan(2x) sec(2x) dx.
Step 5: Next, let's substitute u = sec(2x). This means we need to find du in terms of dx.
Step 6: Differentiate both sides of the equation u = sec(2x) with respect to x:
du/dx = 2sec(2x)tan(2x).
Solving for dx, we get dx = du / (2sec(2x)tan(2x)).
Step 7: Substitute these values into the integral:
∫ tan^2 (2x) sec^4 (2x) tan(2x) sec(2x) dx
becomes
∫ tan^2 (2x) sec^4 (2x) tan(2x) sec(2x) (du / (2sec(2x)tan(2x)))
= 1/2 ∫ tan^2 (2x) sec^4 (2x) du.
Step 8: Simplify the integrand: tan^2 (2x) sec^4 (2x).
Step 9: Now our integral becomes 1/2 ∫ (sec^2 (2x) - 1) sec^4 (2x) du.
Step 10: Expand and distribute: 1/2 ∫ (sec^6 (2x) - sec^4 (2x)) du.
Step 11: Integrate term by term: 1/2 ∫ sec^6 (2x) du - 1/2 ∫ sec^4 (2x) du.
Step 12: Evaluate the indefinite integrals of sec^6 (2x) and sec^4 (2x). This might require using integration formulas or trigonometric identities specific to these integral functions.
Step 13: Finally, simplify your answer by plugging in u = sec(2x) back into the equation. Remember to include the constant of integration, which is denoted as + c.
Based on these steps, please recheck your calculation and ensure that you've evaluated each integral correctly.