Question

find out integral of (secx)^3 dx

Answer 1

Let's asume secx=a and tanx=b
so da=sec²x=secx X tanx dx

by using integration by parts, we get
$secx tanx - \int\limits^a_b { tan^{2} }secx \, dx$

$secx tanx - \int\limits^a_b { (sec^{2}-1) }secx \, dx$

$secx tanx - \int\limits^a_b { (sec^{3}x-secx) } \, dx$

$secx tanx - \int\limits^a_b { sec^{3}x } \, dx + \int\limits^a_b {sec x} \, dx$

$\int\limits^a_b { sec^{3}x } \, dx = secx tanx - \int\limits^a_b { sec^{3}x } \, dx + \int\limits^a_b {sec x} \, dx$

$\int\limits^a_b { sec^{3}x } \, dx + \int\limits^a_b { sec^{3}x } \, dx = secx tanx + \int\limits^a_b {sec x} \, dx$

$2\int\limits^a_b { sec^{3}x } \, dx = secx tanx + \int\limits^a_b {sec x} \, dx$

$2\int\limits^a_b { sec^{3}x } \, dx = \frac{secx tanx + \int\limits^a_b {sec x} \, dx }{y}$

$\int\limits^a_b { sec^{3}x } \, dx = \frac{secx tanx + \int\limits^a_b {sec x} \, dx }{2}$

$\int\limits^a_b { sec^{3}x } \, dx = \frac{secx tanx + ln(secx+tanx)}{y}$$So, the answer is \frac{secx tanx + ln(secx+tanx)}{2} + C$

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Answer 2

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