# Hello. I am having trouble with this calc problem. Integrate, with as few steps as needed, `int tan^3(5x)dx` ` `

embizze | Certified Educator

`int tan^3(5x)dx`

Note that tan^3(5x)=tan(5x)[tan^2(5x)]

=tan(5x)[sec^2(5x)-1] using a trig substitution

=tan(5x)sec^2(5x)-tan(5x)

So the integral can be written as the difference of two integrals:

`int tan(5x)sec^2(5x)dx - int tan(5x)dx`

For the first integral, use a u substitution: Let u=tan(5x), then du=5tan(5x)sec^2(5x). Thus we can rewrite the integrand as 1/5u du. For the second integral, we can integrate tanvdv, so let v=5x and dv=5 so rewrite the integral as 1/5tanv dv.

`1/5 int u du - 1/5 int tan(v) dv`

`=1/5[u^2/2 + C_1 +1/5ln|cosv|+C_2]`

Substituting for u and v we get:

`=1/10[tan^2(5x)+2ln|cos(5x)|]+C`

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Note that in the first integral, we could also let u=sec(5x) and du=5sec(5x)tan(5x). With this substitution the final answer will be:

`1/10[sec^2(5x)+2ln|cos(5x)|]+C`

You can check both answers by differentiating.

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nick-teal | Certified Educator

We can integrate tan^3(5x) by manipulating it with trig properties, and then by using u substitution.

`tan^3(5x) = (sin^3(5x))/(cos^3(5x))`

`(sin^3(5x))/(cos^3(5x)) = sin(5x)(sin^2(5x))/(cos^3(5x))`

So we can rewrite our integral as:

`int (1 - cos^2(5x))/(cos^3(5x))sin(5x)dx`

And now we will use u substitution

Let

`u = cos(5x)`

`du = -5sin(5x)dx`

`(-1/5)du = sin(5x)dx`

Now we sub in u for cos(5x) and (-1/5)du for sin(5x)dx

`int (1-u^2)/u^3(-1/5)du`

Simplify

`(-1/5)intu^-3 - u^-1 du`

Integrate

`(-1/5) (u^-2/-2 - ln(u)) + C`

Sub cos(5x) back in for u

`(-1/5)((cos^-2(5x))/-2 - ln(cos(5x))) = 1/(10cos^2(5x)) + (1/5)ln(cos(5x)) + C`

I know its more than a few simple steps. Hope this helps!!!

Remember that you can also change 1/cos to sec if you want to, so the solution can be written as:

`(1/10)sec^2(5x) + (1/5)ln(cos(5x)) + C`

embizze | Certified Educator

``

Note that tan^3(5x)=tan(5x)[tan^2(5x)]

=tan(5x)[sec^2(5x)-1] using a trig substitution

=tan(5x)sec^2(5x)-tan(5x)

So the integral can be written as the difference of two integrals:

``

For the first integral, use a u substitution: Let u=tan(5x), then du=5tan(5x)sec^2(5x). Thus we can rewrite the integrand as 1/5u du. For the second integral, we can integrate tanvdv, so let v=5x and dv=5 so rewrite the integral as 1/5tanv dv.

``

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Substituting for u and v we get:

``