A lvl H2 Maths: Integration

Prove that cosec^-1 x = sin^-1 (1/x).
Hence find, for x > 0, $\int x cosec^{-1}x dx$

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Answer:

Let y = cosec^-1 x
Thus,
cosec y = x
sin y = 1/x
y = sin^-1 (1/x)

Since y = cosec^-1 x, therefore
cosec^-1 x = sin^-1 (1/x)(proved)

$\int x cosec^{-1}x dx$
$=\int x \sin^{-1}\frac{1}{x} dx$
$=\frac{1}{2}x^{2}\sin^{-1}\frac{1}{x} - \frac{1}{2} \int x^{2}\frac{1}{\sqrt{1-\frac{1}{x^{2}}}}\left ( -\frac{1}{x^{2}} \right ) dx$
$=\frac{1}{2}x^{2}\sin^{-1}\frac{1}{x} + \frac{1}{2} \int \frac{1}{\sqrt{\frac{x^{2}-1}{x^{2}}}} dx$
$=\frac{1}{2}x^{2}\sin^{-1}\frac{1}{x} + \frac{1}{2} \int \frac{x}{\sqrt{x^{2}-1}} dx$
$=\frac{1}{2}x^{2}\sin^{-1}\frac{1}{x} + \frac{1}{4} \int \frac{2x}{\sqrt{x^{2}-1}} dx$
$=\frac{1}{2}x^{2}\sin^{-1}\frac{1}{x} + \frac{1}{4} \frac{\sqrt{x^{2}-1}}{\frac{1}{2}}+c$
$=\frac{1}{2}x^{2}\sin^{-1}\frac{1}{x} + \frac{1}{2} \sqrt{x^{2}-1}+c$

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A lvl H2 Maths: Integration