Integration Techniques (Trigo By Parts)
Could I please get some help on this integration question? My attempts are attached and I would like to know if it is correct. I also think my attempt is rather complicated and would like to know if there is another way to solve the question. Thank you.
Find the antiderivative :
$$\int {(1+\cot \space x)^2 \csc \space x \space dx}$$
*Correction: It should be substitute (ii) into (i) instead of (iii) into (ii)
1 Answer
$\begingroup$$$\int(1+\cot x)^2\csc x\ dx$$
$$=\int\csc\ dx+2\int\csc\cot x\ dx+\int\cot^2x\csc x\ dx$$
The first two are elementary.
$$\int\cot^2x\csc x\ dx=\int\dfrac{\cos^2x}{\sin^3x}\ dx=\int\dfrac{\cos^2x}{\sin^4x}\sin x dx$$
Set $\cos x=u$
Alternatively integrating by parts,
$$\int\cot^2x\csc x\ dx=\int \cos x\dfrac{\cos x}{\sin^3x}\ dx$$
$$=\cos x\int\dfrac{\cos x}{\sin^3x}\ dx-\int\left(\dfrac{d(\cos x)}{dx}\int\dfrac{\cos x}{\sin^3x}\ dx\right)dx$$
$$=-\dfrac{\cos x}{\sin^2x}-\int\dfrac{\sin x}{\sin^2x}\ dx=?$$
$\endgroup$
