What's the integral of #int (tanx)^5*(secx)^4dx#?

Question

Ezekiel Alexander · Accepted Answer

#I=tan^6x/6+tan^8x/8+c# We know that, #color(red)(int[f(x)]^n*f^'(x)dx=[f(x)]^(n+1)/(n+1)+c)# #f(x)=tanx=>f^'(x)=sec^2x# So, #I=int(tanx)^5*(secx)^4dx=int(tanx)^5(sec^2x)(secx)^2dx# #I=int(tanx)^5(1+tan^2x)(sec^2x)dx# #I=int(tanx)^5sec^2xdx+int(tanx)^7sec^2xdx# #I=int(tanx)^5d/(dx)(tanx)dx+int(tanx)^7d/(dx)(tanx)dx# #I=(tanx)^(5+1)/(5+1)+(tanx)^(7+1)/(7+1)+c# #I=tan^6x/6+tan^8x/8+c#

Paisley Johnson · Answer

undefined To solve the integral ∫(tan(x))^5(sec(x))^4 dx, we can use trigonometric identities and integration by substitution. We start by expressing sec(x) as 1/cos(x) and tan(x) as sin(x)/cos(x). Then, we use the substitution u = cos(x), which implies du = -sin(x) dx. After substituting and simplifying, we obtain the integral in terms of u. Finally, we can integrate with respect to u and then revert back to the variable x.

The integral can be solved using trigonometric identities and integration by substitution. First, express sec(x) as 1/cos(x) and tan(x) as sin(x)/cos(x). Then, use the substitution u = cos(x), implying du = -sin(x) dx. After substituting and simplifying, the integral becomes an integral in terms of u. Finally, integrate with respect to u and revert back to the variable x.