Question

Question #53d06

Answer 1

`inttan^3 xsec^6 x` `dx=1/24(3tan^8x+8tan^6x+6tan^4x)+"C"`

Explanation:

`inttan^3 xsec^6 x` `dx`

Let `u=tanx`

Then `du=sec^2x` `dx`

And `dx=cos^2x` `du`

Also `sec^2x=tan^2 x+1=u^2+1`

`inttan^3 xsec^6 x` `dx=intu^3sec^6x xxcos^2x` `du=intu^3sec^4x` `du`

`=int u^3(u^2+1)^2` `du`

Expansion of the bracket yields:

`intu^7+2x^5+u^3` `du=1/8u^8+1/3u^6+1/4u^4+"C"`

`=1/8tan^8x+1/3tan^6x+1/4tan^4x+"C"`

`=1/24(3tan^8x+8tan^6x+6tan^4x)+"C"`

Answer 2

Here we want to save a factor of `sec^2(x)`.

This is so we can apply the standard result for the integral of `"f"(x)*["f"(x)]^n.` I will derive this result here.

Consider,

`"I" = \int "f"'(x)*["f"(x)]^n "d"x`

Substitute `u = "f"(x)`. Then `("d"u)/("d"x)="f"'(x)`, `"d"x = 1/("f"'(x)) "d"u`.

`"I"` is transformed to,

`"I" = \int "f"'(x) * [u]^n * 1/("f"'(x)) "d"u`
`"I" = \int u^n "d"u`.
`"I" = u^(n+1)/(n+1) + C`.

We conclude,

`\int "f"'(x)["f"(x)]^n "d"x = (["f"(x)]^(n+1))/(n+1)+C`.

If `"f"(x) = tan(x)`, `"f"'(x)=sec^2(x)`.

Then,

`\int sec^2(x)tan^n(x)=tan^(n+1)(x)/(n+1)+C` *.

Then we can integrate any power of tan(x) multiplied by `sec^2(x)`.

If we consider your specific integral, `\int sec^6(x)tan^3(x) "d"x`,
we can rewrite it as

`\int sec^2(x) * sec^4(x) * tan^3(x) "d"x`.

We know that `sec^2(x) = 1 + tan^2(x)`. Then,

`sec^4(x) = (1+tan^2(x))^4`
`sec^4(x) = tan^8(x) + 4tan^6(x) + 6tan^4(x) + 4tan^2(x) + 1`.

Then,

`\int sec^2(x) (tan^8(x) + 4tan^6(x) + 6tan^4(x) + 4tan^2(x) + 1) tan^3(x) "d"x = \int sec^2(x)tan^11(x) "d"x + 4\int sec^2(x)tan^9(x) "d"x + 6\int sec^2(x)tan^7(x) "d"x + 4\int sec^2(x)tan^5(x) "d"x + \int sec^2(x)tan^3(x) "d"x`.

Using * from earlier,

`\int sec^6(x)tan^3(x) "d"x = (tan^12(x))/(12) + (2tan^10(x))/(5)+(3tan^8(x))/4+(2tan^6(x))/3+(tan^4(x))/4+C`

Answer 3

The integral may be found by substitution. Either `u = tanx` or `u = secx` will work. Here I've used `u = secx`.

Explanation:

For choice of `u` see the note at the end.

`int tan^3xsec^6x dx = int tan^2xsec^5x (secxtanx) dx`

`= int (sec^2 x -1) sec^5 x (secxtanx) dx`

`= int (sec^7 x - sec^5 x) (secxtanx) dx`

With `u = sec x` so `du = secxtanx dx`, this is

`= int (u^7-u^5) du`

`= u^8/8-u^6/6+C`

Reversing the substitution, we finish with

`= 1/24sec^6x(3sec^2x-4) +C`

Note

It is instructive to compare this solution with that given by Monzur R.

When there is a choice of substitutions, I expect it to be simpler to keep the trig function that appears to the higher power and use trig identities to replace the function that has the lower power.

For `int tan^3xsec^6x dx`, I've used

`int tan^3xsec^6x dx = int tan^2xsec^5x (secxtanx) dx` and `u = secx`

to get `int(u^2-1)u^5 du`

If the exponents were reversed, I 'd have made the other choice.

If we had `int tan^5xsec^4x dx`, I should have used

`int tan^5xsec^4x dx = int tan^5xsec^2x (sec^2x) dx` and `u = tanx`

to get `int u^5(u^2+1) du`

Answer 4

You are free to do either. I consider one choice simpler than the other.

Explanation:

`inttan^3 x sec^6 x dx`

We can evaluate this integral using either of two substitutions. (And probably some other ways as well.)

Your book's first method
If we use the fact that `d/dx(tanx) = sec x`, we get

`inttan^3 x sec^6 x dx = int tan^3 x sec^4 x (sec^2xdx)`

`= int tan^3x(tan^2x+1)^2 du`

`= int u^3(u^2+1)^2 du`.

Finish by expanding the integrand and integrating.

`= int (u^7+2u^5+u^3) du`.

`= u^8/8+u^6/3+u^4/4+C`

`= tan^8x/8+tan^6x/3+tan^4x/4+C`

`= 1/24(3tan^8x+8tan^6x+6tan^4x)+C`

Your book's second method
Use the fact that `d/dx(secx) = secxtanx`.

`inttan^3 x sec^6 x dx = int tan^2 x sec^5 x (secxtanxdx)`

`= int (sec^2x-1)sec^5x du`

`= int (u^2-1)u^5 du`.

In this case, we do not have the square of a binomial, so the integrand is simpler to expand.

`= int (u^7 - u^5) du`.

`= u^8/8-u^6/6+C`

`= 1/24(3sec^8x-4sec^6x)+C`

The difference between `1/24(3tan^8x+8tan^6x+6tan^4x)` and `1/24(3sec^8x-4sec^6x)` is a constant (which means the two ways of presenting the answer have different `C`'s.)

Note
I suggest rewriting in terms of the function that has the greater exponent. It simplifies the step in which we expand the polynomial integrand.

I would use your book's first method for `int tan^9x sec^4xdx` to get `u^9(u^2+1) du`.

I would not want to use the first method for `int tan^3x sec^8x dx` because with `u=tanx`, we'd need to expand `u^3(u^2+1)^3 du`.
The second method uses `u=secx` to get `(u^2-1)u^7 du`.