Question

Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle.\(\)

\(\int {\frac{{dx}}{{{x^2}\sqrt {4 - {x^2}} }},x = 2sin\theta } \)

Short answer

The integration \(\int {\frac{{dx}}{{{x^2}\sqrt {4 - {x^2}} }}} \) after substituting \(x = 2sin\theta \) is \( - \frac{1}{{4x}}\sqrt {4 - {x^2}} + c\), and the right-angled triangle will be

Step-by-step solution

Substituting \(x = 2sin\theta \) in the integration.

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

And we have to substitute.

\(x = 2sin\theta \,\,\,....\left( 2 \right)\)

Differentiating both sides with respect to x.

\(\begin{aligned}{l}\frac{d}{{dx}}.x &= \frac{d}{{dx}}\left( {2sin\theta } \right)\\1 &= 2\frac{d}{{dx}}sin\theta \\1 &= 2\,cos\theta .\frac{{d\theta }}{{dx}}\\dx &= 2\,cos.d\theta \end{aligned}\)

Putting the value of \(x\) and \(dx\) in 1, we have

\(\int {\frac{{2cos\theta \,d\theta }}{{{{\left( {2sin\theta } \right)}^2}\sqrt {4 - {{\left( {2sin\theta } \right)}^2}} }}} \)

\(\begin{aligned}{l} &= \int {\frac{{2cos\theta \,d\theta }}{{4si{n^2}\theta \sqrt {4 - 4si{n^2}\theta } }}} \\ &= \int {\frac{{2cos\theta \,d\theta }}{{4si{n^2}\theta \,\,2\sqrt {1 - si{n^2}\theta } }}} \\ &= \int {\frac{{cos\theta .d\theta }}{{4si{n^2}\theta .cos\theta }}} \\ &= \frac{1}{4}\int {\frac{{d\theta }}{{si{n^2}\theta }}} \\ &= \frac{1}{4}\int {cose{c^2}\theta \,\,d\theta } \\ &= \frac{1}{4}\left( { - cot\theta } \right) + c\end{aligned}\)

Finding the sides of the triangle.

So, \( = \frac{1}{4}\left( { - cot\theta } \right) + c\)

But, \(x = 2 sin\theta \)

So, \(sin\theta = \frac{x}{2}\)

And

\(\begin{aligned}{l}cot\,\theta &= \frac{{\sqrt {1 - si{n^2}\theta } }}{{sin\theta }}\\ &= \frac{{\sqrt {1 - \frac{{{x^2}}}{4}} }}{{\frac{x}{2}}}\\ &= \frac{{\sqrt {4 - {x^2}} }}{x}\end{aligned}\)

So, \( - \frac{1}{4}cot\theta + c\)

\( - \frac{1}{4}\frac{{\sqrt {4 - {x^2}} }}{x} + c\)

Now, \(sin\,\theta = \frac{{perpendicular}}{{hypoteneus}}\)

And also \(sin \theta = \frac{x}{2}\)

So, \(p = x,h = 2\) and

By Pythagoras theorem,

\(\begin{aligned}{l}{b^2} + {p^2} &= {h^2}\\{b^2} + {x^2} &= {\left( 2 \right)^2}\\{b^2} &= 4 - {x^2}\\b &= \sqrt {4 - {x^2}} \end{aligned}\)

Building right angle triangle.

Since, \(h = 2, p = x\,\,and b = \sqrt {4 - {x^2}} \)

So, the triangle will be

Hence, the integration \(\int {\frac{{dx}}{{{x^2}\sqrt {4 - {x^2}} }}} \) after substituting \(x = 2 sin\theta \) will be \( - \frac{1}{4}\frac{{\sqrt {4 - {x^2}} }}{x} + c\)and the triangle will be