\(\int{x}^{3}\sqrt{{x}^{2}-1}dx\)
I think I am supposed to use integration by substitution or integration by parts, but I'm not sure. Can anyone help with this? Thanks :)
+26387 I think I am supposed to use integration by substitution or integration by parts, but I'm not sure. Can anyone help with this?
\(\int{x}^{3}\sqrt{{x}^{2}-1}dx\)
We apply twice substitution
We need the following formulae:
\(\begin{array}{|rcll|} \hline \cosh^2(z) - \sinh^2(z) &=& 1 \\ \sinh^2(z) &=& \cosh^2(z) -1 \\ \sinh(z) &=& \sqrt{\cosh^2(z) -1} \\\\ \cosh^2(z) - \sinh^2(z) &=& 1 \\ \cosh^2(z) &=& 1+\sinh^2(z) \\\\ \frac{d}{dz}\sinh(z) &=& \cosh(z) \\ \frac{d}{dz}\cosh(z) &=& \sinh(z) \\ \hline \end{array} \)
1. Substitution
\(\begin{array}{|rcll|} \hline x &=& \cosh(z)\\ dx &=& \sinh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ \cosh^3(z)\cdot \sqrt{\cosh^2(z)-1} \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh(z) \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ \hline \end{array} \)
2. Substitution
\(\begin{array}{|rcll|} \hline u &=& \sinh(z)\\ du &=& \cosh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ &=& \int{ \cosh^2(z)\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ [1+\sinh^2(z)]\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ (1+u^2)\cdot u^2 \ du } \\ &=& \int{ (u^2+u^4)\ du } \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array} \)
3. Back substitution
\(\begin{array}{|rcll|} \hline u &=& \sinh(z) \\ &=& \sqrt{\cosh^2(z) -1} \\ &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array} \)
+9476 Sorry this got posted 3 times, something was lagging and I couldn't see it so I thought it wasn't posting. 
+9476 I've gotten this far on it but after that I'm stuck:
Set u = x2 and du = 2x dx
\((1/2)\int u\sqrt{u-1} du\)
+129839 x = sec θ dx = sec θ*tan θ dθ
∫ sec^3( θ ) √[ sec^2θ - 1 ] secθ*tanθ dθ =
∫ sec^3( θ ) √ [ tan^2θ ] secθ*tanθ dθ =
∫ sec^3( θ ) √ [ tanθ ] secθ*tanθ dθ =
∫ sec^4 θ * tan^2θ dθ
∫ [sec^2 θ] * [sec^2 θ] * tan^2θ dθ =
∫ tan^2θ * [ tan^2θ + 1] [sec^2 θ] dθ =
∫ [ tan^4θ + tan^2θ ] [sec^2 θ] dθ =
∫ [ tanθ ]^4 * [sec^2 θ] dθ + ∫ [ tanθ ]^2 * [sec^2 θ] dθ =
(1/5) [ tanθ ]^5 + (1/3) [tanθ ]^3 + C
Now x = sec θ ...so.... x^2 = sec^2θ = 1 + [ tanθ]^2 →
x^2 - 1 = [tanθ]^2 ...so......
[x^2 - 1]^(5/2) = [ [tanθ]^2 ] ^(5/2) = [ tanθ ]^5
And
[x^2 - 1]^(3/2) = [ [tanθ]^2 ] ^(3/2) = [ tanθ ]^3
So we have
∫ x^3 √[ x^2 - 1 ] dx =
(1/5)[x^2 - 1]^(5/2) + (1/3)[x^2 - 1]^(3/2) + C
Proof :
http://www.wolframalpha.com/input/?i=derivative++(1%2F5)%5Bx%5E2++-+1%5D%5E(5%2F2)++%2B+(1%2F3)%5Bx%5E2++-+1%5D%5E(3%2F2)+++%2B+C
+26387 I think I am supposed to use integration by substitution or integration by parts, but I'm not sure. Can anyone help with this?
\(\int{x}^{3}\sqrt{{x}^{2}-1}dx\)
We apply twice substitution
We need the following formulae:
\(\begin{array}{|rcll|} \hline \cosh^2(z) - \sinh^2(z) &=& 1 \\ \sinh^2(z) &=& \cosh^2(z) -1 \\ \sinh(z) &=& \sqrt{\cosh^2(z) -1} \\\\ \cosh^2(z) - \sinh^2(z) &=& 1 \\ \cosh^2(z) &=& 1+\sinh^2(z) \\\\ \frac{d}{dz}\sinh(z) &=& \cosh(z) \\ \frac{d}{dz}\cosh(z) &=& \sinh(z) \\ \hline \end{array} \)
1. Substitution
\(\begin{array}{|rcll|} \hline x &=& \cosh(z)\\ dx &=& \sinh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ \cosh^3(z)\cdot \sqrt{\cosh^2(z)-1} \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh(z) \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ \hline \end{array} \)
2. Substitution
\(\begin{array}{|rcll|} \hline u &=& \sinh(z)\\ du &=& \cosh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ &=& \int{ \cosh^2(z)\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ [1+\sinh^2(z)]\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ (1+u^2)\cdot u^2 \ du } \\ &=& \int{ (u^2+u^4)\ du } \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array} \)
3. Back substitution
\(\begin{array}{|rcll|} \hline u &=& \sinh(z) \\ &=& \sqrt{\cosh^2(z) -1} \\ &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array} \)
+26387 I think I am supposed to use integration by substitution or integration by parts, but I'm not sure.
Can anyone help with this?
\(\int{ x^3\cdot\sqrt{x^2-1}\ dx } \)
short way:
Substitution
\(\begin{array}{|rcll|} \hline u &=& \sqrt{x^2-1}=(x^2-1)^{\frac12} \quad & \text{or} \quad x = \sqrt{u^2+1} \\ du &=& \frac12 \cdot (x^2-1)^{-\frac12} \cdot 2x\ dx \\ &=& \frac{x}{\sqrt{x^2-1}}\ dx \quad & \text{or} \quad dx = \frac{\sqrt{x^2-1}}{x}\ du\\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ ( \sqrt{u^2+1})^3\cdot u \frac{\sqrt{x^2-1}}{x}\ du } \\ &=& \int{ ( \sqrt{u^2+1})^3\cdot u \frac{u}{\sqrt{u^2+1}}\ du } \\ &=& \int{ ( \sqrt{u^2+1})^2\cdot u^2\ du }\\ &=& \int{ (u^2+1)\cdot u^2\ du }\\ &=& \int{ (u^4+u^2)\ du }\\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array} \)
Back substitution
\(\begin{array}{|rcll|} \hline u &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array} \)
