Integrals by parts and substitution

**rocsfinest762** · March 15th 2009, 03:20 PM

I'm not understanding what to do to the first problem after I cancel out the cscx in the top and bottom. The 2nd and 3rd problems are still kind of confusing.

1) Int (cscx cotx)/((csc(x))^5) dx

2) Int (dx)/(sqrt(1-16x^2)

3) Int ((x^3)-(3x^2)+3x)/(x-1) dx

**skeeter** · March 15th 2009, 04:38 PM

Originally Posted by rocsfinest762

I'm not understanding what to do to the first problem after I cancel out the cscx in the top and bottom. The 2nd and 3rd problems are still kind of confusing.

1) Int (cscx cotx)/((csc(x))^5) dx

let $u = \csc{x}$ ... $du = -csc{x}\cot{x} \, dx$

2) Int (dx)/(sqrt(1-16x^2)

integral is an arcsin ...

$\frac{1}{4} \int \frac{4}{\sqrt{1 - (4x)^2}} \, dx$

3) Int ((x^3)-(3x^2)+3x)/(x-1) dx

do the long division

.

**Grandad** · March 15th 2009, 11:16 PM

Hello rocsfinest762

Originally Posted by rocsfinest762

...1) Int (cscx cotx)/((csc(x))^5) dx...

$I = \int\frac{\csc x \cot x}{\csc^5x}\,dx$

Now $\csc x \cot x = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}=\frac{\cos x}{\sin^2x}$

and $\csc^5x = \frac{1}{\sin^5x}$

$\Rightarrow I = \int\frac{\cos x\sin^5x}{\sin^2x}\,dx$

$= \int\cos x\sin^3x\,dx$

Let $\sin x = u$ . Then $\cos x \,dx =du$

I think you can complete it now.

Grandad

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