Finding Integrals by Substitution - High School Math

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Question

Evaluate:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }; ; dx$

Answer

Set $u = \cos x$ .

Then $\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \cos x= -\sin x$ and $\sin x ; dx = - du$ .

Also, since $u = \cos x$ , the limits of integration change to $\cos \frac{\pi }{2} = 0$ and $\cos 0 = 1$ .

Substitute:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }; ; dx$

$= \int_{1}^{0} \left (- \sqrt{u} \right ) ; du$

$= \int_{1}^{0} \left (-u ^{\frac{1}{2}} \right ) ; du$

$= \int_{0}^{1} u ^{\frac{1}{2}} ; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \ & \end{matrix}\right|$ $\begin{matrix} 1\ \ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \ \frac{ 2u \sqrt{u}} {3} & \ & \end{matrix}\right|$ $\begin{matrix} 1\ \ 0 \end{matrix}$

$=\frac{ 2\cdot 1\cdot \sqrt{1}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} = \frac{ 2} {3}$

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Question

Evaluate:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}; ; dx$

Answer

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Also, since $u = \ln x$ , the limits of integration change to $\ln e^{3} = 3$ and $\ln 1= 0$ .

Substitute:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}; ; dx$

$\dpi{120} = \int_{0}^{3}\sqrt{u} ; du$

$= \int_{0}^{3} u ^{\frac{1}{2}} ; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \ & \end{matrix}\right|$ $\begin{matrix} 3\ \ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \ \frac{ 2u \sqrt{u}} {3} & \ & \end{matrix}\right|$ $\begin{matrix} 3\ \ 0 \end{matrix}$

$=\frac{ 2\cdot 3\cdot \sqrt{3}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} =2\sqrt{3}$

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Question

Determine the indefinite integral:

$\dpi{120} \int\cos x \sqrt{\sin x }; ; dx$

Answer

Set $u = \sin x$ . Then

$\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \sin x= \cos x$ .

and

$\cos x ; dx = du$

The integral becomes:

$\dpi{120} \int\cos x \sqrt{\sin x }; ; dx$

$\dpi{120} =\int \sqrt{u }; ; du$

$= \int u ^{\frac{1}{2}} ; du$

$\dpi{120} \dpi{150} = \frac{u^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} + C$

$\dpi{150} = \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C$

$\dpi{120} =\frac{ 2u \sqrt{u}} {3} +C$

Substitute back:

$\dpi{120} =\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

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Question

Determine the indefinite integral:

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}; ; dx$

Answer

$\log_{10} x = \frac{\ln x}{\ln 10}$ , so this can be rewritten as

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}; ; dx$

$\dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }; ; dx$

$\dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }; ; dx$

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Substitute:

$\dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }; ; dx$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} ; du$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} ; du$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$= \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$= \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

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Question

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=$

Answer

This integral will require a u-substitution.

Let $u=\sqrt{x}$ .

Then, differentiating both sides, $du=\frac{1}{2\sqrt{x}}dx$ .

We need to solve for dx in order to replace all x terms with u terms.

$dx=(2\sqrt{x})du$ .

$\int\frac{\sqrt{x}}{1-x}\cdot(2\sqrt{x})du=\int\frac{2xdu}{1-x}$

This is a little tricky because we stilll have x and u terms mixed together. We need to go back to our original substitution.

$u=\sqrt{x}\Rightarrow x=u^2$

$\int\frac{2xdu}{1-x}=\int\frac{2u^2 du}{1-u^2}$

Now we have an integral that looks more manageable. First, however, we can't forget about the bounds of the definite integral. We were asked to evaluate the integral from to . Because $u=\sqrt{x}$ , the bounds will change to $\sqrt{4}=2$ and $\sqrt{9}=3$ .

Essentially, we have made the following transformation:

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=\int_{2}^{3}\frac{2u^2}{1-u^2}du$ .

The latter integral is easier to evaluate.

$\int_{2}^{3}\frac{2u^2}{1-u^2}du=-2\int_{2}^{3}\frac{-u^2}{1-u^2}du$

$=-2\int_{2}^{3}\frac{1-u^2-1}{1-u^2}du=-2\int_{2}^{3}\left ( \frac{1-u^2}{1-u^2}+\frac{-1}{1-u^2} \right )du$

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

At this point, we can separate the integral into two smaller integrals.

$-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du=-2\int_{2}^{3}1\cdot du+2\int_{2}^{3}\frac{1}{1-u^2}du$ .

The integral $-2\int_{2}^{3}1\cdot du$ evaluates to -2, so now we just need to worry about the other integral. This will require the use of partial fraction decomposition. We need to rewrite $\frac{1}{1-u^2}$ as the sum of two fractions.

$\frac{1}{1-u^2}=\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$

We need to solve for the values of A and B.

$\frac{A}{1-u}+\frac{B}{1+u}=\frac{1}{(1-u)(1+u)}$

This means that and . This is a relatively simple system of equations to solve, so I won't go into detail. The end result is that $A=B=\frac{1}2{}$ .

$\frac{1}{1-u^2}=\frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u}$

Let's now go back to the integral $2\int_{2}^{3}\frac{1}{1-u^2}du$ .

$2\int_{2}^{3}\frac{1}{1-u^2}du=2\int_{2}^{3}\left ( \frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u} \right )du$

Distribute the 2 to both integrals and separate it into two integrals.

$=\int_{2}^{3}\frac{1}{1-u}du + \int_{2}^{3}\frac{1}{1+u}du$

$=-\ln|1-u| ]_{2}^{3} + \ln|1+u|]_{2}^{3}$

$=-\ln|-2|+\ln|-1|+\ln4-\ln3$

$=-\ln2+\ln1+\ln4-\ln3$

$=\ln\frac{2}{3}$ .

Remember we need to add this value back to the value of $-2\int_{2}^{3}1\cdot du$ , which we already determined to be -2.

The final answer is $-2+\ln\frac{2}{3}$ .

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