Method
We consider the following cases:
1) n even let u = tanx and use the following identity: sec²x = tan²x + 1
2) m odd u = secx tan²x = sec²x -1
2) m even n odd reduce power of secx: tan²x = sec²x -1
Example I
Calculate ∫tan²xsec⁴xdx
sec⁴x is the near derivative of tan²x and n is even we write u = tanx.
Taking the derivative of both sides we have du = sec²xdx. dx = du/sec²x.
Let's substitute tanx and dx
∫tan²xsec⁴xdx = ∫u²sec⁴x.du/sec²x
= ∫u²sec²xdu
We use the identity sec²x = tan²x + 1 in order to have the integral as a function of u. Then by substituting tanx by u: sec²x = u² + 1. Therefore:
∫tan²xsec⁴xdx = ∫u²( u² + 1)du
= ∫(u⁴ + u²)du
= u⁵/5 + u³ + C
= 1/5tan⁵x + 1/3tan³x + C
Example II
Solve ∫tan³xsec³xdx.
We start by looking at the condition for n. Here n is odd. There isn't a condition for n only. We look at the condition for m. Here m is odd then we write u = secx and tan²x = sec²x-1
Let's write u = secx. Then du = tanxsecxdx. Therefore dx = du/tanxsecx.
Let's substitute secx and dx.
∫tan³xsec³xdx = ∫tan³xu³.du/tanxu
= ∫tan²xu²du
Let's express tanx in function of secx so that we can have the integral as a function of u
∫tan³xsec³xdx = ∫(sec²x - 1)u²du
= ∫(u² - 1)u²du
= ∫u⁴du - ∫u²du
= 1/5u⁵ - 1/3u³ + C
= 1/5sec⁵x - 1/3sec³x + C
Practice
Solve:
1) ∫tan²xsec³x
2) ∫tan³xsec⁴x
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