A) State the chain rule for integration
Ans. The chain rule for integration is also known as " Integration by substitution "
Integration by substitution is taken in order to make integration solve easily in few steps.
For, [tex]I = \int\limits (x+2)^{2} \,dx[/tex]
Instead of expanding term [tex](x+2)^{2}[/tex]
With substitution of [tex]u = (x+2) [/tex] and [tex]du= 1 dx [/tex]
We simplified the integration as
[tex]I = \int\limits (u)^{2} \, du[/tex]
[tex]I = \frac{(u)^{3}}{3}+C[/tex]
By replacaing value of u=x+2
[tex]I = \frac{(x+2)^{3}}{3}+C[/tex]
B) State the rule of differentiation for the sine function.
Ans. We know that [tex]\frac{d}{dx}Sinx dx = Cosx [/tex]
C) Find the indefinite integral using substitution.
Ans.
Given, [tex]I = \int\limits {\frac{Cos14x}{Sin14x} } \, dx[/tex]
Take y = Sin14x
Differentiating both side
[tex]dy=14Cos14x dx [/tex]
[tex]\frac{dy}{14} = Cos14x\, dx[/tex]
Substituting values in integration,
[tex]I = \int\limits {\frac{Cos14x}{Sin14x} } \, dx[/tex]
[tex]I = \int\limits {\frac{1}{y} } \,\frac{dy}{14} [/tex]
[tex]I = \frac{1}{14}\int\limits {\frac{1}{y} } \,dy [/tex]
[tex]I = \frac{1}{14} lny + C [/tex]
Replacing values in the integration
[tex]I = \frac{1}{14} ln(14Sin14x) + C [/tex]
D)Check your work by taking a derivative of your answer from part C.
Ans.
Answer for Part C is [tex]I = \frac{1}{14} ln(14Sin14x) + C [/tex]
Differentiating the answer
we get,
[tex]=\frac{1}{14}\frac{d}{dx}[ ln(Sin14x) + C]\\=\frac{1}{14}\frac{1}{Sin14x} \frac{d}{dx}(Sin14x)+ \frac{d}{dx}C\\=\frac{1}{14}\frac{1}{Cos14x}(14Cos14x)\\=\frac{Cos14x}{Sin14x} \\ =I[/tex]
