How do I integrate tanx (1+sec^4 x)^3/2 dx
My daughter is doing surface area problems, and all the examples assume she knows how to finish it off once we get to the integration....
All the online integration calculators I tried said that this cannot be integrated.
Recheck your problem.
Try substitution:
u=cos(x)
du=-sin(x)dx
Then
∫tanx (1+sec^4 x)^3/2 dx
=∫(1+sec^4 x)^3/2 sin(x) dx / cos(x)
=-∫[(1+(1/u)^4)^3/2 /u] du
But the integration is still ugly!
Rechecking the problem is the first thing to do.
To integrate the given function, let's break it down step by step.
First, let's simplify the integrand by using trigonometric identities. We can rewrite sec^4(x) as (sec^2(x))^2, and then use the identity sec^2(x) = tan^2(x) + 1 to replace sec^2(x).
tan(x) * (1 + sec^4(x))^3/2 dx
= tan(x) * (1 + (tan^2(x) + 1)^2)^3/2 dx
= tan(x) * (1 + tan^4(x) + 2tan^2(x) + 1)^3/2 dx
= tan(x) * (2tan^4(x) + 2tan^2(x) + 2)^3/2 dx
Next, we can use a substitution to simplify the integrand further. Let's substitute u = tan(x) so that du = sec^2(x) dx.
Now, the integrand becomes:
u * (2u^4 + 2u^2 + 2)^3/2 du
To integrate this, we can use the power rule for integration. The power rule states that the integral of x^n is (x^(n+1))/(n+1). Applying this rule to each term in the integrand, we get:
∫ (2u^6 + 2u^4 + 2u^2)^(3/2) du
Integrating each term separately, we have:
∫ (2u^6)^(3/2) du + ∫ (2u^4)^(3/2) du + ∫ (2u^2)^(3/2) du
Simplifying further, we have:
2∫ u^9 du + 2∫ u^6 du + 2∫ u^3 du
Using the power rule, we integrate each term:
2 * (u^(9+1))/(9+1) + 2 * (u^(6+1))/(6+1) + 2 * (u^(3+1))/(3+1)
Simplifying this further, we have:
(1/5)u^10 + (2/7)u^7 + (1/2)u^4
Finally, substitute back u = tan(x) into the expression to obtain the final antiderivative:
(1/5)tan^10(x) + (2/7)tan^7(x) + (1/2)tan^4(x) + C
Thus, integrating tan(x) * (1 + sec^4(x))^3/2 dx gives us the expression (1/5)tan^10(x) + (2/7)tan^7(x) + (1/2)tan^4(x) + C, where C is the constant of integration.