Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 31

int tan^5 (x) dx
To solve, apply the Pythagorean identity tan^2(x) = sec^2(x) - 1 until the integrand is in the form int u^n du .
= int tan^3(x) tan^2(x) dx
= int tan^3(x)(sec^2x-1)dx
= int [tan^3(x) sec^2(x) - tan^3(x)]dx
= int [tan^3(x)sec^2(x) - tan(x) tan^2(x)]dx
= int [tan^3(x) sec^2(x) - tan(x) (sec^2x-1)]dx
= int [tan^3(x) sec^2(x) - tan(x)sec^2(x) +tan(x)]dx
= int tan^3(x)sec^2(x)dx - int tan(x) sec^2(x) dx + int tan(x) dx
For the first and second integral, apply u-substitution method. Let u be:
u = tan x
Then, differentiate u.
du= sec^2(dx)
Plugging them, the first and second integral becomes:
= int u^3 du - int u du + int tan (x) dx
Then, apply the integral formula int x^n dx = x^(n+1)/(n+1) + C and int tan(theta)d theta = ln |sec (theta)| + C .
= u^4/4 - u^2/2 + ln |sec(x)| + C
And, substitute back u = tan(x).
= (tan^4(x))/4 - (tan^2(x))/2 + ln|sec(x)| + C

Therefore, int tan^5(x) dx= (tan^4(x))/4 - (tan^2(x))/2 + ln|sec(x)| + C .

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