Question

Question #4323f

Answer 1

`int tg(x)dx = tG(x)+c, cinRR`, where `G(x)` is an antiderivative of `g(x)`. Whether a substitution is needed depends on `g`'s formula.

Explanation:

Assuming `t` is a constant with respect to `x`, all you have to do is find an antiderivative of `g(x)`. If we call that antiderivative `G(x)`, then you have

`int tg(x)dx = t int g(x)dx = tG(x)+c.`

Answer 2

`int\ tan(x)\ dx=-ln|cos(x)|+C`

Explanation:

I assume that you want to find the integral of the tangent function, or `int\ tan(x)\ dx`.

First, using trigonometric identities, rewrite it to
`int\ sin(x)/cos(x)\ dx`

Now, we need to use u-substitution. For this, it would be best if we could find some expression in the integrand that can be substituted to a single variable and which derivative, when divided into the integrand, can simplify so that no `x`'s occur.

In this example, we will choose `u=cos(x)`, because the integrand divided by its derivative `(du)/dx=-sin(x)` will simplify dramatically.

Then, we have
`int\ sin(x)/cos(x)dx/(du)\ du=-int\ sin(x)/u1/sin(x)\ du=-int\ (du)/u`.

This is simply
`-ln|u|+C`.

Substitute back `u=cos(x)` to get the final answer of
`-ln|cos(x)|+C`.

Answer 3

`int "tg"x dx = ln|secx|+"c"`

Explanation:

Firstly, we shall let `"tg"-=tan`

So our integral is `inttanxdx=int(secxtanxdx)/secx`

Now let `u=secx` and `du=secxtanxdx`. We then transform the integral to get it as `int1/udu = ln|u|=ln|secx|+"c"`