Question

How do you evaluate the integral of `int (2^tanθ)sec^2(θ) dθ`?

Answer 1

`int 2^{tan theta} * sec^2 theta "d"theta = 2^{tan theta}/ln2 + "Integration Constant"`

Explanation:

Note that the

`frac{"d"}{"d"theta}(tan theta) = sec^2 theta`

The integrand looks to be a result of the chain rule so let's work backwards.

The derivative of an exponential function is retains the exponential. So we begin by finding the derivative of `2^{tan theta}`.

`frac{"d"}{"d"theta}(2^{tan theta}) = frac{"d"}{"d"theta}(e^{ln2 * tan theta})`

`= e^(ln2 * tan theta) * "d"/("d"theta)(ln2 * tan theta)`

`= e^(ln2 * tan theta) * ln2 * sec^2 theta`

`= 2^{tan theta} * ln2 * sec^2 theta`

Therefore,

`int 2^{tan theta} * ln2 * sec^2 theta "d"theta = 2^{tan theta} + "Constant"`

Divide both sides by `ln2` to get our final answer.

`int 2^{tan theta} * sec^2 theta "d"theta = 2^{tan theta}/ln2 + "Another Constant"`