1. ## help meeeeee:(:(:(((

1)INT dx/(cosx+sinx)^4 ????

2)INT dx/(sin^3x+cos^3x) ????

3)INT dx/(sinx+2secx)^2 ?????

4)INT dx/[(sin^4x)cosx] ????

5)INT [(cosx + sinx)/(sin2x^1/2)]dx ??????

2. Originally Posted by berkanatci
1)INT dx/(cosx+sinx)^4 ????

2)INT dx/(sin^3x+cos^3x) ????

3)INT dx/(sinx+2secx)^2 ?????

4)INT dx/[(sin^4x)cosx] ????

5)INT [(cosx + sinx)/(sin2x^1/2)]dx ??????
INT 1/(cosx + sinx)^4 dx

Multiply out the denominator:
INT 1/(cos^4x + 4cos^3x*sinx + 6cos^2x*sin^2x + 4cosx*sin^3x + sin^4x) dx

Now, group terms to reduce:
INT 1/[(cos^4x + 2cos^2x*sin^2x + sin^4x) + (4cos^3x*sinx + 4cosx*sin^3x) + 4cos^2x*sin^2x] dx
INT 1/[(cos^2x + sin^2x)^2 + 4cosx*sinx(cos^2x + sin^2x) + 4cos^2x*sin^2x] dx
INT 1/(1 + 4cosx*sinx + 4cos^2x*sin^2x) dx

When factored, the denominator becomes:
INT 1/(2cosx*sinx +1)^2 dx
INT 1/(sin2x + 1)^2 dx

Multiply numerator and denominator by (1 - sin2x)^2
INT (1 - sin2x)^2/(1 - sin^2(2x))^2 dx
INT (1 - 2sin2x + sin^2(2x))/cos^4(2x) dx

Let 1 = sin^2(2x) + cos^2(2x)
INT (cos^2(2x) - 2sin2x + 2sin^2(2x))/cos^4(2x) dx

Separate the fraction:
INT cos^2(2x)/cos^4(2x) dx - INT 2sin(2x)/cos^4(2x) dx + INT 2sin^2(2x)/cos^4(2x) dx
INT sec^2(2x) dx - 2*INT sec^2(2x)[tan2x*sec2x] dx + 2*INT tan^2(2x)[sec^2(2x)] dx

First:
INT sec^2(2x) dx = 1/2*tan2x

Second:
INT sec^2(2x)[tan2x*sec2x] dx

Let u = sec2x <--> du = 2*tan2x*sec2x dx --> dx = 1/(2tan2x*sec2x) du
1/2*INT u^2 du
1/6*u^3
1/6*sec^3(2x)

Third:
INT tan^2(2x)[sec^2(2x)] dx

Let u = tan2x <--> du = 2*sec^2(2x) dx --> dx = 1/(2sec^2(2x)) du
1/2*INT u^2 du
1/6*u^3
1/6*tan^3(2x)

Putting these back to gether, we get:
(
1/2*tan2x) - 2(1/6*sec^3(2x)) + 2(1/6*tan^3(2x)) + C

1/2*tan2x - 1/3*sec^3(2x) + 1/3*tan^3(2x) + C

3. OMG, I would be making some good money if I were actually paid to do this stuff.

4. These questions so easy berkan.
You must work too much...

5. Originally Posted by pinsim
These questions so easy berkan.
I wouldn't say that

Look at http://www.mathhelpforum.com/math-he...ease-help.html

At least, the second one is very easy, rather than berkanatci's problems.