Math Is Fun Forum

Hastega

Member

Registered: 2007-06-20

Posts: 7

Help! Integral Calculus

Can someone help me with these two problems?

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Help! Integral Calculus

According to Wikipedia table of integrals:

but it didnt have anything for cotangent.

---

The Beginning Of All Things To End.
The End Of All Things To Come.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Help! Integral Calculus

Possibly split the cotangent into sine and cosine, then use integration by parts?  Just a guess.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

HallsofIvy

Guest

Re: Help! Integral Calculus

The first is an easy "standard" problem because it has sine to an ODD power.

  Write it as int sin^6(x) (sin(x)dx) = int (1- cos^2(x)^3 (sin(x)dx).  Now let u= cos(x) so that du= -sin(x)dx and we have -int (1- u^2)^3 du= -int (1- 3u^2+ 3u^4- u^6)du= -(u- u^3+ (3/5)u^5- (1/7)u^7)+ C= (1/7)cos^7(x)-(3/5)cos^5(x)+ cos^3(x)- cos(x)+ C.

  I'm surprised Wikipedia doesn't have such a formula for cot.  My Calculus book gives
int cot^n (x)dx= - (cot^(n-1)(x))/(n-1)- int cot^(n-2)(x)dx.

Hastega

Member

Registered: 2007-06-20

Posts: 7

Re: Help! Integral Calculus

So (1/7)cos^7(x)-(3/5)cos^5(x)+ cos^3(x)- cos(x)+ C is the answer to number 1.

I believe the formula for Cot is

But i still can't answer it, I always get stuck because there not integrable.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Help! Integral Calculus

for 1 i get:

which according wolfram integrator can be simplified to:

Last edited by luca-deltodesco (2007-08-15 20:33:36)

---

The Beginning Of All Things To End.
The End Of All Things To Come.

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: Help! Integral Calculus

Problem 2:

To solve this we need only the basic result:

Which is easily obtained via the trig relation:

For then:

for n >= 2.

(Which is the formula metioned by HallsofIvy)

So we can always reduce to the case n = 1 or n = 0.

This will allow you to solve the given integral.

Last edited by gnitsuk (2007-08-20 20:57:54)