Friday

August 29, 2014

August 29, 2014

Posted by **rox** on Sunday, June 24, 2012 at 2:12pm.

∫tan^6 (x) dx

- calculus -
**Count Iblis**, Sunday, June 24, 2012 at 4:35pmPut tan(x) = u ---->

x = arctan(u) ---->

dx = du/(1+u^2)

Integral of u^6/(1+u^2) du =

Integral of

(u^6 + u^4 - u^4 - u^2 + u^2 + 1 -1)/

(u^2 + 1) du =

Integral of

(u^4 - u^2 + 1) du - Integral of

1/(1+u^2) du =

1/5 u^5 - 1/3 u^3 + u - arctan(u) + c=

1/5 tan^5(x) - 1/3 tan^3(x) + tan(x) - x + c

**Related Questions**

calculus (check my work please) - Not sure if it is right, I have check with the...

Calculus - Question - Am I allowed to do this? for the integral of ∫ sec^4...

Calculus - Calculate the following integral: ∫ sec^4 (3x)/ tan^3 (3x) dx ...

Calculus - Find the integral by substitution ∫ [(16 x3)/(x4 + 5)] dx &#...

Calculus AP - I'm doing trigonometric integrals i wanted to know im doing step ...

Calculus 2 correction - I just wanted to see if my answer if correct the ...

Calculus - Evaluate the following definite integral: integral at a = -1, b=2 -...

COLLEGE CALCULUS. HELP! - Evaluate the definite integral ∫(0,2) (x-1)^25 ...

Calculus - Evaluate the triple integral ∫∫∫_E (x+y)dV where E ...

Calculus - Evaluate the triple integral ∫∫∫_E (x)dV where E is...