integrate ∫(2cos3x + 3sinx) / sin^3x dx and ∫tan^5x csc^5x dx

integrate ∫(2cos3x + 3sinx) / sin^3x dx

$$\small{\begin{array}{l|lll} \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx \quad & \quad \cos{(3x)}= \cos{(x)}\cos{(2x)} -\sin(x)\sin{(2x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}[1-2\sin^2{(x)}]-\sin{(x)}\cdot 2\sin{(x)}\cos{(x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}[1-2\sin^2{(x)}]-2\sin^2{(x)}\cos{(x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}-2\cos{(x)}\sin^2{(x)}-2\sin^2{(x)}\cos{(x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}-4\cos{(x)}\sin^2{(x)}\\ =\int \dfrac{2[\cos{(x)}-4\cos{(x)}\sin^2{(x)}] + 3\sin{(x)} }{ \sin^3{(x)} }\ dx & \\ =\int \dfrac{2\cos{(x)}-8\cos{(x)}\sin^2{(x)} + 3\sin{(x)} }{ \sin^3{(x)} }\ dx & \\ = 2 \int \dfrac{ \cos{(x)} } { \sin^3{(x)} }\ dx -8 \int \dfrac{ \cos{(x)}\sin^2{(x)} } { \sin^3{(x)} }\ dx +3 \int \dfrac{ \sin{(x)} } { \sin^3{(x)} }\ dx & \\ = 2 \int \dfrac{ 1 } { \sin^2{(x)} }\cdot\cot{(x)}\ dx -8 \int \dfrac{ \cos{(x)} } { \sin{(x)} }\ dx +3 \int \dfrac{ 1 } { \sin^2{(x)} }\ dx &\\ \quad & \quad \text{Formula:} \\ \quad & \quad \boxed{ \int \frac{f'(x)}{f(x)}=\ln{(f(x))} ~~ \int \frac{\cos{(x)}}{\sin{(x)}}\ dx = \ln {( \sin{(x)} )} } \\ \quad & \quad \boxed{ (\cot{(x)})' = -\frac{1}{\sin^2{(x)}} ~~\int \frac{1}{\sin^2{(x)}}\ dx = -\cot{(x)}}\\ \quad & \quad \boxed{ \int f'(x) \cdot [f(x)]^1 = \frac{ [f(x)]^2}{2} ~~\int \dfrac{ 1 } { \sin^2{(x)} }\cdot\cot{(x)}\ dx = -\frac{\cot^2{(x)}}{2} }\\ = 2 (-\frac{\cot^2{x}}{2}) - 8 \ln {( \sin{(x)} )} + 3 (-\cot{(x)} )&\\\\ = -\cot^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1&\\ \end{array}}\\\\\\$$

$$\small{\begin{array}{rcl} \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& -\cot^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1 \quad | \quad -\cot^2{(x)}=1-\csc^2{(x)}\\\\ \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& 1-\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1 \\\\ \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& -\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + (c_1+1) \quad | \quad c = c_1+1 \\\\ \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& -\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c \\\\ \end{array}}$$

hi Bhustancrowe

Mmm

I spent ages doing it the wrong way, here is the right way.  I can't take the credit.  :(

Oh I managed to delete the bit i did correctly - this is the second one - you should

be able to do the first little bit of it  yourself   

Now it is right     I had just made a careless error right at the beginning.

Thanks for spotting it Alan     [ I do have a question at the bottom though]

the correct answer is   

I have not found my errors  :(

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$$\\\int\frac{(2cos3x + 3sinx) }{sin^3x }dx\\\\ =\int\frac{2(cos2xcosx-sin2xsinx) + 3sinx }{sin^3x }dx\\\\ =\int\frac{2cos2xcosx - 2sin2xsinx) + 3sinx }{sin^3x }dx\\\\ =\int\frac{2(cos^2x-sin^2x)cosx - 2(2sinxcosx)sinx) + 3sinx }{sin^3x }dx\\\\ =\int\frac{2cosxcos^2x-2cosxsin^2x - 4sin^2xcosx + 3sinx }{sin^3x }dx\\\\ =\int\frac{2cos^3x-2cosxsin^2x - 4sin^2xcosx + 3sinx }{sin^3x }dx\\\\ =\int\frac{2cos^3x - 6sin^2xcosx + 3sinx }{sin^3x }dx\\\\ =\int\;2cot^3x-\frac{6cosx}{sinx} + 3csc^2x \;dx\\\\ =\int\;2cot^3x\;dx -\int\;\frac{6cosx}{sinx}\;dx +\int\; 3csc^2x \;dx\\\\ =\int\;2cot^3x\;dx -6ln(sinx) +\int\; 3csc^2x \;dx\\\\ =-6ln(sinx)\;+\;\int\;2cot^3x\;dx -3cotx\\\\ =-6ln(sinx)\;-3cotx\;+\;\int\;2cot^3x\;dx$$

$$\\$NOW I'll use the reduction formula$\\\\ \int\;2cot^3x\;dx\\\\ =2[\frac{-Cot^{3-1}x}{3-1}-\int\;cot^{-2+3}x\;dx]\\\\ =2[\frac{-Cot^{2}x}{2}-\int\;cotx\;dx]\\\\ =-Cot^{2}x-2\;ln(sinx)\\\\$$

$$\\$So - continuing from before$\\\\ =-6ln(sinx)\;-3cotx\;+\;\int\;2cot^3x\;dx \\\\ =-6ln(sinx)\;-3cotx\;+\; -Cot^{2}x-2\;ln(sinx) +c \\\\ =-8ln(sinx)\;-3Cotx-Cot^{2}x \;+c \\\\$$

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Now this is correct but Wolfram Alpha is telling me that   $$cot^2x=csc^2x$$    for the restricted values involed here.

WHY IS THAT ?

Melody, 

$$\\ \sin^2+\cos^2=1 \text{ Divide through by }\sin^2 \text{ to get }1+\cot^2=\csc^2$$

So cot² differs from csc² by 1.  This 1 is wrapped up in the constant of integration.  i.e. your constant of integration differs from Wolfram Alpha's by 1.

. 

Thanks Alan,       I should have thought of that    

Thanks Heureka :)

that you very much guys , im a little confused but i'll try to understand it(im kinda slow in math)