# calculus

posted by .

find the indefinite integral:
integration of tan^3(7x) dx

• calculus -

Yo can find a reduction formula for the integral of tan^n(x) as follows. We have that

tan^n(x) = sin^n(x)/cos^n(x) =

sin^(n-2)(x)/cos^n(x) sin^2(x) =

sin^(n-2)(x)/cos^n(x) [1-cos^2(x)] =

sin^(n-2)(x)/cos^n(x) -
sin^(n-2)(x)/cos^(n-2)(x) =

sin^(n-2)(x)/cos^(n-2)(x) 1/cos^2(x) -
sin^(n-2)(x)/cos^(n-2)(x) =

tan^(n-2)(x) 1/cos^2(x) - tan^(n-2)(x)

Now, 1/cos^2(x) is the derivate of
tan(x), so you can immediately integrate the first term:

Integral of tan^(n-2)(x) 1/cos^2(x)dx =

1/(n-1) tan^(n-1)(x)

The integral of the second term is, of course, a similar problem as the original problem, but with a lower value for n, so you can interate the formula until you end up at n = 1 or n = 0.

So, denoting the integral of tan^n(x)dx by I_n, we have:

I_n = 1/(n-1) tan^(n-1)(x) - I_{n-2}

For n = 3, we get:

I_3 = 1/2 tan^2(x) - I_1

and I_1 is the integral of tan(x)dx, which is -Log|cos(x)|.

## Similar Questions

1. ### Calculus

Would someone clarify this for me... Is antiderivatives just another name for intergral and why is intergral of a function is the area under the curve?
2. ### Calculus

Solve these indefinite and definite integrals. [integration sign] 4j-5j^3 dj I got 2j^2 - 5/4j^4... is this my final answer?
3. ### calculus

How do I derive the integration reduction formula for tangent?
4. ### calculus

Use integration by parts to evaluate the integral of x*sec^2(3x). My answer is ([x*tan(3x)]/3)-[ln(sec(3x))/9] but it's incorrect. u=x dv=sec^2(3x)dx du=dx v=(1/3)tan(3x) [xtan(3x)]/3 - integral of(1/3)tan(3x)dx - (1/3)[ln(sec(3x))/3] …
5. ### Calculus

"Evaluate the following indefinite integral using integration by parts: *integral sign* tan^-1(x) dx" I let u = tan^-1(x) and dv = dx. Is that right?
6. ### calculus

Using an integration formula,what is the indefinite integral of (sign for integral)(cos(4x)+2x^2)(sin(4x)-x)dx. Any help very much appreciated.
7. ### integration (calculus)

Find the indefinite integral. ∫((e^-x-1)/(e^-x + x)^2) dx I am trying to use the substitution method. u=e^-x + x du= -e^-x+1 dx Right so far?
8. ### Calc BC

1. Find the indefinite integral. Indefinite integral tan^3(pix/7)sec^2(pix/7)dx 2. Find the indefinite integral by making the substitution x=3tan(theta). Indefinite integral x*sqrt(9+x^2)dx 3. Find the indefinite integral. Indefinite …