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With Worked Solution![The following rules may he applied \\ $\begin{aligned} & \displaystyle \int \dfrac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+c \\ & \displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1} \dfrac{x}{a}+c \\ & \displaystyle \int \dfrac{-1}{\sqrt{1-x^2}} d x=\cos ^{-1} x+c \\ & \displaystyle \int \dfrac{-1}{\sqrt{a^2-x^2}} d x=\cos ^{-1} \dfrac{x}{a}+c \\ & \displaystyle \int \dfrac{1}{1+x^2} d x=\tan ^{-1} x+c \\ & \displaystyle \int \dfrac{1}{a^2+x^2} d x=\dfrac{1}{a} \tan ^{-1} \dfrac{x}{a}+c . \end{aligned}$\\ \begin{multicols}{2} \textbf{Example 1}\\ %question 123221 Evaluate \(\displaystyle \int_{-1}^1 \dfrac{1}{\sqrt{2-x^2}}\, d x\) \\ \textbf{Example 1 solution}\\ $\begin{aligned} \displaystyle \int_{-1}^1 \frac{1}{\sqrt{2-x^2}} d x &=\int_{-1}^1 \frac{1}{\sqrt{(\sqrt{2})^2-x^2}} d x \\ &=\left[\sin ^{-1} \frac{x}{\sqrt{2}}\right]_{-1}^1 \\ &=\left[\sin ^{-1} \frac{1}{\sqrt{2}}-\sin ^{-1} \frac{-1}{\sqrt{2}}\right] \\ &=\left[\sin ^{-1} \frac{1}{\sqrt{2}}+\sin ^{-1} \frac{1}{\sqrt{2}}\right] \\ &=2 \sin ^{-1} \frac{1}{\sqrt{2}} \\ &=2 \times \frac{\pi}{4} \\ &=\frac{\pi}{2} \end{aligned}$\\ \columnbreak \textbf{Example 2}\\ %question 123223 Evaluate \(\displaystyle \int_0^{\frac{1}{3}} \dfrac{1}{1+9 x^2}\, d x\)\\ \textbf{Example 2 solution}\\ $\begin{aligned} \displaystyle \int_0^{\frac{1}{3}} \frac{1}{1+9 x^2} d x &=\int_0^{\frac{1}{3}} \frac{1}{9\left(\frac{1}{9}+x^2\right)} d x \\ &=\frac{1}{9} \int_0^{\frac{1}{3}} \frac{1}{\left(\frac{1}{3}\right)^2+x^2} d x \\ &=\frac{1}{9}\left[3 \tan ^{-1} 3 x\right]_0^{\frac{1}{3}} \\ &=\frac{1}{3}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\ &=\frac{1}{3} \times \frac{\pi}{4} \\ &=\frac{\pi}{12} \end{aligned}$\\ \end{multicols}](/media/azuffa4b/integrations-that-give-inverse-trig-functions.png)
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For Example: Find \(\int {\dfrac{1}{{\sqrt {4 - {x^2}} }}} \,dx\)\( = {\sin ^{ - 1}}\dfrac{x}{2} + c\)
For Example: Find \(\int {\dfrac{1}{{\sqrt {1 - 4{x^2}} }}} \,dx\)
\(\begin{align*}\int {\dfrac{1}{{\sqrt {1 - 4{x^2}} }}} \,dx &= \int {\dfrac{1}{{\sqrt {4\left( {{{\left( {1/2} \right)}^2} - {x^2}} \right)} }}} \,dx\\ &= \dfrac{1}{2}\int {\dfrac{1}{{\sqrt {{{\left( {1/2} \right)}^2} - {x^2}} }}} \,dx = \dfrac{1}{2}{\sin ^{ - 1}}2x\,\, + c\end{align*}\)
For Example: Find \(\int {\dfrac{1}{{9 + {x^2}}}} \,dx\)
\(\int {\dfrac{1}{{9 + {x^2}}}} \,dx = \dfrac{1}{3}{\tan ^{ - 1}}\dfrac{x}{3} + c\)
For Example: Find \(\int {\dfrac{1}{{1 + 9{x^2}}}} \,dx\)
\(\begin{align*}\int {\dfrac{1}{{1 + 9{x^2}}}} \,dx &= \int {\dfrac{1}{{9\left( {{{\left( {1/3} \right)}^2} + {x^2}} \right)}}} \,dx\\ &= \dfrac{1}{9} \times 3{\tan ^{ - 1}}3x\, = \dfrac{1}{3}{\tan ^{ - 1}}3x\,\, + c\end{align*}\)