Re: Integration of inverse Trig functions
paulmdrdo said:
Please tell me if i worked out the problem correctly.
1. ∫(dx/x2-x+2)
completing the square of the denominator i have,
∫[dx/(x-1)2+12]
a=1, u=x-1; du=dx
∫(du/a2+u2)
1/1*tan-1x-1/1 + c = tan-1x-1 + C -->> final answer
2. ∫[dx/(15+2x-x2)1/2]
∫[dx/(14-(x-1)2)1/2]
∫{dx/([(14)1/2]2-(x-1)2)}
a = (14)1/2; u = x-1; du = dx
= 1/(14)1/2*sin-1x-1/(14)1/2 + C --->>final answer
and P.S how to use that font that you are using?
Since the OP has gotten the correct answer to Q1 and is very close to getting the correct answer to Q2, I will post how I would approach these problems, which involves trigonometric or hyperbolic substitutions.
Q1.
[math]\displaystyle \begin{align*} \int{ \frac{dx}{x^2 - x + 2}} &= \int{ \frac{dx}{ \left( x - \frac{1}{2} \right) ^2 + \frac{7}{4} } } \end{align*}[/math]
Now make the substitution [math]\displaystyle \begin{align*} x - \frac{1}{2} = \frac{\sqrt{7}}{2}\tan{(\theta)} \implies dx = \frac{\sqrt{7}}{2}\sec^2{(\theta)}\,d\theta \end{align*}[/math] and the integral becomes
[math]\displaystyle \begin{align*} \int{ \frac{dx}{ \left( x - \frac{1}{2} \right) ^2 + \frac{7}{4} } } &= \int{ \frac{\frac{\sqrt{7}}{2}\sec^2{(\theta)}\,d\theta}{ \left[ \frac{\sqrt{7}}{2}\tan{(\theta)} \right] ^2 + \frac{7}{4} } } \\ &= \frac{\sqrt{7}}{2} \int{ \frac{\sec^2{(\theta)}\,d\theta}{\frac{7}{4}\tan^2{(\theta)} + \frac{7}{4}} } \\ &= \frac{\sqrt{7}}{2}\cdot \frac{4}{7} \int{\frac{\sec^2{(\theta)}\,d\theta}{\tan^2{( \theta )} + 1}} \\ &= \frac{2\sqrt{7}}{7} \int{ \frac{\sec^2{(\theta)}\,d\theta}{\sec^2{(\theta)}} } \\ &= \frac{2\sqrt{7}}{7}\int{1\,d\theta} \\ &= \frac{2\sqrt{7}}{7} \theta + C \\ &= \frac{2\sqrt{7}}{7}\arctan{\left[ \frac{2\sqrt{7}}{7}\left( x - \frac{1}{2} \right) \right] } + C \end{align*}[/math]
Q2.
[math]\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{15 + 2x - x^2}}} &= \int{\frac{dx}{\sqrt{-\left( x^2 - 2x - 15 \right) } } } \\ &= \int{ \frac{dx}{\sqrt{-\left[ x^2 - 2x + (-1) ^2 - (-1)^2 - 15 \right] }}} \\ &= \int{ \frac{dx}{\sqrt{- \left[ (x - 1)^2 - 16 \right] }}} \\ &= \int{ \frac{dx}{\sqrt{16 - (x - 1)^2}}} \end{align*}[/math]
Now make the substitution [math]\displaystyle \begin{align*} x - 1 = 4\sin{(\theta)} \implies dx = 4\cos{(\theta)}\,d\theta \end{align*}[/math] and the integral becomes
[math]\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{16 - (x-1)^2}}} &= \int{\frac{4\cos{(\theta)}\,d\theta}{\sqrt{16 - \left[ 4\sin{(\theta)} \right] ^2 }}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{16 - 16\sin^2{(\theta)}}}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{16 \left[ 1 - \sin^2{(\theta)} \right] }}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{16\cos^2{(\theta)}}}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{4\cos{(\theta)}}} \\ &= \int{ 1\,d\theta} \\ &= \theta + C \\ &= \arcsin{\left[ \frac{1}{4} \left( x - 1 \right) \right] } + C \end{align*}[/math]