# Trig substitution help

1. Feb 2, 2008

1. The problem statement, all variables and given/known data

$$\int$$ ((sin(x))^3/(cos(x)) )*dx

3. The attempt at a solution

alright i have been trying to use

u= cosx
-du = sinx

but it dosnt make sense bause there is still a sinx^2 to account for

so i know i need to make a trig substitution but i cant figure out the appropriate
substitution for sinx^2 even though i think the 1/2 ( 1- cos2x) is hte only one

question... when u use a u substituion for u = cosx -du = sindx would u write

1/2 $$\int$$ ((sinx*( 1 - cos2x)) / cosx)*dx

- 1/2 $$\int$$( (1 - 2u) / u ) *du

if u = cosx and u have to change a cos2x... what would you write 2u... that wouldn't make sensebecause it would come out 2*cosx which does not equatl cos2x

idk where im lost... i cant logically follow through this problem for some reason, any advice would be amazing, thanks for your time.

Last edited: Feb 2, 2008
2. Feb 2, 2008

### Mystic998

Assuming your work is right, just write (1 - 2u)/u = 1/u - 2

3. Feb 2, 2008

according the calculator ti 89 the answer is

-ln abs(cosx) - sinx^2 / 2

accordign to what i ended up w/ i got

-1/2 ( ln abs(cosx) - 2*cosx )

which dosnt make sense my work is wrong it has to be lol

if u have a few mins to try to work through it on paper let me kno what u see

4. Feb 2, 2008

### Mystic998

Oops, I misread. Just a second.

Edit: Your work is wrong. cos(2x) =/= 2u if u = cos(x).

I'd suggest instead of using $sin^{2}(x) = \frac{1 - cos(2x)}{2}$ that you use a more basic identity.

Last edited: Feb 2, 2008
5. Feb 2, 2008

i know how to evaluate those easy integrals but the challenge is turning a complex integral into a easy integral

6. Feb 2, 2008

1/u du is

ln(u)

2 =2u

u = cosx

7. Feb 2, 2008

yea i have been trying to figure out a simpler substitution and it just isnt working

8. Feb 2, 2008

### Mystic998

What's the very first trig identity you ever learned?

9. Feb 2, 2008

### Littlepig

what about using $$sin(x)^{2}=1-cos(x)^{2}$$ ?

you get 2 simple integrals i guess...

10. Feb 2, 2008

sinx^2 + cosx^2 = 1

11. Feb 2, 2008

omg i made that way to hard >.< lol thanks
ill reevaluate it and see what happens

12. Feb 3, 2008

### fermio

$$\int\frac{\sin^3 x}{\cos x}dx=\int\frac{\sin x\frac{1-\cos(2x)}{2}}{\cos x}dx=\frac{1}{2}\int\frac{\sin x(1-\cos(2x))}{\cos x}\frac{d(\cos x)}{-\sin x}=-\frac{1}{2}\int\frac{1-\cos(2x)}{\cos x}d(\cos x)=$$
$$=-\frac{1}{2}\int\frac{1-(2\cos^2 x-1)}{\cos x}d(\cos x)=-\frac{1}{2}\int\frac{2}{\cos x}-2\cos x d(\cos x)=-\ln\cos x+\frac{\cos^2 x}{2}+C$$

where $$d(\cos x)=-\sin x dx$$
$$dx=\frac{d(\cos x)}{-\sin x}$$
$$\cos(2x)=2\cos^2 x-1$$

Answer should be $$-\ln\cos x+\frac{1}{4}\cos (2x)+C$$ according to http://integrals.wolfram.com/index.jsp

$$\int\frac{\sin^3 x}{\cos x}dx=\int\frac{\sin x(1-\cos^2 x)}{\cos x}dx=\int\frac{\sin x(1-\cos^2 x)}{\cos x}\frac{d(\cos x)}{-\sin x}=-\int\frac{1-\cos^2 x}{\cos x}d(\cos x)=$$
$$=-\int\frac{1}{\cos x}-\cos x d(\cos x)=-(\ln\cos x-\frac{\cos^2 x}{2})+C$$
where $$d(\cos x)=-\sin x dx$$
$$dx=\frac{d(\cos x)}{-\sin x}$$

Does this http://integrals.wolfram.com/index.jsp integrator giving wrong answer? Becouse $$\frac{\cos^2 x}{2}$$ not equal to $$\frac{1}{4}\cos (2x)$$

Last edited: Feb 3, 2008
13. Feb 3, 2008

### Big-T

$$\frac{\cos^2(x)}{2}=\frac{\cos(2x)}{4}+C_1$$

14. Feb 3, 2008

### fermio

Interesting
$$\frac{\cos^2 0.5}{2}\approx 0.385075576$$
$$\frac{\cos 1}{4}\approx 0.135075576$$
$$\frac{\cos^2 0.5}{2}-\frac{\cos 1}{4}=0.25$$
$$\frac{\cos^2 0.3}{2}-\frac{\cos 0.6}{4}=0.25$$
...
But why?

15. Feb 3, 2008

### awvvu

cos^2(x) = 1/2(1 + cos(2x))

16. Feb 3, 2008

$$\int$$ 1 / (sinx - 1) dx

wth i cant figure this out, thinkin about it to hard again

i broke a more complex integral down into this final part, diffrent problem

i know it equals

cosx / (sinx - 1)

17. Feb 3, 2008

### Big-T

Try conjugation.

18. Feb 3, 2008

reduces back into itself

19. Feb 3, 2008