1. ## ∫sin3x/sin2x dx

Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.

2. Originally Posted by gillyjay
Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.
$\displaystyle \frac{sin 3x}{sin 2x}=\frac{3sin x-4sin^3 x}{2sin xcos x}=\frac{3-4sin^2 x}{2cos x}=\frac{3-4(1-cos^2x)}{2cosx}=\frac{4cos^2x-1}{2cosx}=2cosx-\frac{secx}{2}$

Can you finish it?

3. Originally Posted by gillyjay
Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.
Numerator: sin(3x) = sin(2x) cos(x) + sin(x) cos(2x).

Denominator: sin(2x) = 2 sin(x) cos(x).

Substitute the above, break up the integral into two parts and simplify. The following will be useful: integrate Cos&#91;2x&#93;&#47;Cos&#91;x&#93; - Wolfram|Alpha

Clcik on Show steps.

4. A probably somewhat an overkill substitution would be to let:
$x = 2\arctan{t}$ then find $-\int \frac{(t^2-3)(3t^2-1)}{(t-1)(t+1)(t^2+1)^2}\;{dt}$.

5. Originally Posted by gillyjay
Hi guys I have a problem getting anywhere on this integral, could need some help how to solve it.
$\displaystyle\int{\frac{Sin3x}{Sin2x}}dx=\int{\fra c{Sin2xCosx+Cos2xSinx}{Sin2x}}dx=\int{Cosx}dx+\int {\frac{Cos2xSinx}{2CosxSinx}dx$

$=\displaystyle\int{Cosx}dx+\int{\frac{2Cos^2x-1}{2Cosx}}dx=\int{Cosx}dx+\int{Cosx}dx-\frac{1}{2}\int{Secx}dx$

6. Thanks for the hints guys, just a Q, archie, How did you evaluate Sin3x into Sin2xCosx+Cos2xSinx?

Edit: This problem is indeed eating my mind, how do we go from here to get and to evaluate to get integral value of: 2 sinx + 1/4 ln[1-sinx] / [1 + sinx ] + c (The Right answer)

7. Originally Posted by gillyjay
Thanks for the hints guys, just a Q, archie, How did you evaluate Sin3x into Sin2xCosx+Cos2xSinx?
Apply the identity $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\ sin\beta$ with $\alpha=2x$ and $\beta=x$ (or the other way around...you'll still get the same result).

8. Thanks alot!, hmm but I don't really get the last part on how to get the answer

9. Originally Posted by gillyjay
Thanks alot!, hmm but I don't really get the last part on how to get the answer
Do you mean, on arriving at

$\displaystyle\ 2\int{Cosx}dx-\frac{1}{2}\int{Secx}dx$

at which point

$\displaystyle\int{Cosx}dx=Sinx+C$

and

$\displaystyle\int{Secx}dx=ln|Secx+Tanx|+C$

or before that ?

The other identity used in my previous post was

$\displaystyle\ Cos^2x=\frac{1}{2}\left(1+Cos2x\right)\Rightarrow\ 2Cos^2x-1=Cos2x$

10. If you're wondering how to evaluate $\displaystyle \int{\sec{x}\,dx}$...

$\displaystyle \int{\sec{x}\,dx} = \int{\frac{1}{\cos{x}}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{\cos^2{x}}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{1 - \sin^2{x}}\,dx}$

$\displaystyle = \int{\frac{1}{1 - u^2}\,du}$ after making the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$

$\displaystyle = \int{\frac{1}{(1 - u)(1 + u)}\,du}$

$\displaystyle = \int{\frac{1}{2(1 - u)} + \frac{1}{2(1 + u)}\,du}$

$\displaystyle = -\frac{1}{2}\ln{|1 - u|} + \frac{1}{2}\ln{|1 + u|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{1 + u}{1 - u}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{1 + \sin{x}}{1 - \sin{x}}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{(1 - \sin{x})(1 + \sin{x})}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{1 - \sin^2{x}}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{\cos^2{x}}\right|} + C$

$\displaystyle = \frac{1}{2}\ln{\left|\left(\frac{1 + \sin{x}}{\cos{x}}\right)^2\right|} + C$

$\displaystyle = \ln{\left|\left[\left(\frac{1+\sin{x}}{\cos{x}}\right)^2\right]^{\frac{1}{2}}\right|} +C$

$\displaystyle = \ln{\left|\frac{1 + \sin{x}}{\cos{x}}\right|} + C$

$\displaystyle = \ln{\left|\frac{1}{\cos{x}} + \frac{\sin{x}}{\cos{x}}\right|} + C$

$\displaystyle = \ln{|\sec{x} + \tan{x}|} + C$.

11. Originally Posted by gillyjay

This problem is indeed eating my mind, how do we go from here to get and to evaluate to get integral value of: 2 sinx + 1/4 ln[1-sinx] / [1 + sinx ] + c (The Right answer)

Also notice that

$\displaystyle\ ln{\left|secx+tanx\right|}=ln{\left|\frac{1}{cosx} +\frac{sinx}{cosx}\right|}=ln{\left|\frac{1+sinx}{ cosx}\right|}$

$=\displaystyle\frac{1}{2}ln\left[\frac{1+sinx}{cosx}\right]^2=\frac{1}{2}ln\left[\frac{(1+sinx)^2}{cos^2x}\right]$

$=\displaystyle\frac{1}{2}ln\left[\frac{(1+sinx)(1+sinx)}{1-sin^2x}\right]=\frac{1}{2}ln\left[\frac{(1+sinx)(1+sinx)}{(1-sinx)(1+sinx)}\right]$

$=\displaystyle\frac{1}{2}ln{\left|\frac{1+sinx}{1-sinx}\right|}=-\frac{1}{2}ln{\left|\frac{1-sinx}{1+sinx}\right|}$

Multiplying that by a half gives the result in the form you are looking for.

12. Originally Posted by gillyjay
Thanks alot!, hmm but I don't really get the last part on how to get the answer
Please show what you have done and say where you get stuck.

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# integrate sin2x.sin3x

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