# Tangent substitution in trigonometric substitution

Find:

$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}}\;dx$$

Let $u=2x$ and with tangent substitution we have $x=\frac{3}{2}\,tan\,\theta$ and now we have $dx=\frac{3}{2}\,sec^2\,d\theta$.

Also, $\sqrt{4x^2+9}=3\,sec\,\theta$.

When $x=0$, $tan\,\theta=0$, so $\theta=0$; when $x=3\sqrt{3}/2$, $tan\,\theta=\sqrt{3}$ so $\theta=\pi/3$

$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}}\,dx=\int^{\pi/3}_0\frac{\frac{27}{8}\,tan^3\,\theta}{27\,sec^3\,\theta}\,\frac{3}{2}\,sec^2\,\theta\,d\theta$$ $$=\frac{3}{16}\int^{\pi/3}_0\frac{1-cos^2\,\theta}{cos^2\,\theta}sin\,\theta\,d\theta$$

With u substitution $u=cos\,\theta$ and $du=-sin\,\theta\,d\theta$. When $\theta=0$, $u=1$; when $\theta=\pi/3$, $u=\frac{1}{2}$

Therefore: $$\frac{3}{16}\int^{\pi/3}_0\frac{1-cos^2\,\theta}{cos^2\,\theta}sin\,\theta\,d\theta=\frac{-3}{16}\int^{\frac{1}{2}}_1\frac{1-u^2}{u^2}du$$

My question lies in the next part:

$$=\frac{3}{16}\int^{\frac{1}{2}}_1(1-u^{-2})du$$

Why does $\frac{-3}{16}$ become positive? Also what did they do to get $1-u^{-2}$?

• On your second-to-last line, you have the integral from 1 to 3/2; should be 1 to 1/2. – mr_e_man Feb 9 '18 at 18:35

$$\frac{1-u^2}{u^2}=\frac{1}{u^2}-\frac{u^2}{u^2}=u^{-2}-1=-(1-u^{-2})$$
$$\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \frac { 1-u^{ 2 } }{ u^{ 2 } } du=\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( \frac { 1 }{ { u }^{ 2 } } -1 \right) du=\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( { u }^{ -2 }-1 \right) du=\\ =\frac { 3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( 1-{ u }^{ -2 } \right) du$$