Question

Question #a06e9

Answer 1

`intsec(x+a)dx=ln(abs(sec(x+a)+tan(x+a)))+C`

Explanation:

You may know the straightaway rule:

`intsec(t)dt=ln(abs(sec(t)+tan(t)))+C`

In case you don't, we can derive this rule and solve your problem simultaneously.

`intsec(x+a)dx`

This is a difficult move to foresee, which is why it's likely a good move to memorize the secant integral. But, we can do the integration by multiplying the numerator and denominator by `sec(x+a)+tan(x+a)`:

`=intsec(x+a)(sec(x+a)+tan(x+a))/(sec(x+a)+tan(x+a))dx`

`=int(sec^2(x+a)+sec(x+a)tan(x+a))/(sec(x+a)+tan(x+a))dx`

Now, let `u=sec(x+a)+tan(x+a)`. Its derivative is: `du=(sec(x+a)tan(x+a)+sec^2(x+a))dx`.

Note that `u` is the denominator and `du` is the numerator:

`=int(du)/u`

Which is the natural logarithm integral:

`=ln(absu)+C`

So:

`=ln(abs(sec(x+a)+tan(x+a)))+C`

Which follows the rule `intsec(t)dt=ln(abs(sec(t)+tan(t)))+C` if you, from the outset, let `t=x+a` implying that `dt=dx`.