∫ tan^2 (x) sec^4 (x) dx
∫ [tan^2 (t) + tan^4 (t)] dt
∫ [1-tan^2 (x)] / [sec^2 (x)] dx
Trigonometric integral
Please show steps so I can understand!
tan^2 (sec^4) = tan^2(1+tan^2)sec^2
= (tan^2 + tan^4)(sec^2) dx
Now note that if u = tanx, you have
(u^2 + u^4) du
See what you can do with the others.
I got the first two, but I'm unsure with the last one.
I got the answer to be -ln |secxtanx|+C
Here's what I did:
∫ [1-tan^2 (x)] / [sec^2 (x)] dx
∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx)
∫cosx-sinx(sinx/cosx)
∫cosx-∫sin^2(x)/cosx
sinx-∫(1-cos^2(x))/cosx
sinx-∫(1/cosx)-cosx
sinx-∫secx-∫cosx
sinx-sinx-∫secx
=-ln |secxtanx|+C
Can someone just verify that I did everything correctly?
To solve the integral ∫ tan^2 (x) sec^4 (x) dx, we can rewrite it using trigonometric identities and then apply a substitution.
Step 1: Rewrite the integral using trigonometric identities.
Using the identity tan^2 (x) = sec^2 (x) - 1, we can rewrite the integral as:
∫ (sec^2 (x) - 1) sec^4 (x) dx.
Step 2: Simplify the expression.
∫ sec^6 (x) dx - ∫ sec^4 (x) dx.
Step 3: Apply a substitution.
Let's substitute u = sec(x), which means du = sec(x)tan(x) dx.
Step 4: Express the integrals in terms of the new variable.
For the first integral:
∫ sec^6 (x) dx = ∫ u^6 du.
For the second integral:
∫ sec^4 (x) dx = ∫ (sec^2 (x))^2 dx = ∫ u^4 du.
Step 5: Integrate each term.
For the first integral:
∫ u^6 du = u^7/7 + C.
For the second integral:
∫ u^4 du = u^5/5 + C.
Step 6: Substitute back in the original variable x.
For the first integral:
∫ sec^6 (x) dx = sec^7 (x)/7 + C.
For the second integral:
∫ sec^4 (x) dx = sec^5 (x)/5 + C.
Step 7: Combine the two integrals.
∫ sec^6 (x) dx - ∫ sec^4 (x) dx = sec^7 (x)/7 - sec^5 (x)/5 + C.
Therefore, the solution to the integral ∫ tan^2 (x) sec^4 (x) dx is sec^7 (x)/7 - sec^5 (x)/5 + C.