Finding Integrals by Substitution - Math

Card 1 of 20

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Evaluate:

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

Tap to reveal answer

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}$

← Didn't Know|Knew It →

Question

Evaluate:

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

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

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

Tap to reveal answer

Answer

, so this can be rewritten as

$\dpi{120}=\int$\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$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$$

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 as the sum of two fractions.

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}$2{}$ .

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}$$ .

← Didn't Know|Knew It →

Question

Evaluate:

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

Tap to reveal answer

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}$

← Didn't Know|Knew It →

Question

Evaluate:

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

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

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

Tap to reveal answer

Answer

, so this can be rewritten as

$\dpi{120}=\int$\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$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$$

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 as the sum of two fractions.

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}$2{}$ .

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}$$ .

← Didn't Know|Knew It →

Question

Evaluate:

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

Tap to reveal answer

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}$

← Didn't Know|Knew It →

Question

Evaluate:

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

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

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

Tap to reveal answer

Answer

, so this can be rewritten as

$\dpi{120}=\int$\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$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$$

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 as the sum of two fractions.

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}$2{}$ .

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}$$ .

← Didn't Know|Knew It →

Question

Evaluate:

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

Tap to reveal answer

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}$

← Didn't Know|Knew It →

Question

Evaluate:

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

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

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

Determine the indefinite integral:

Tap to reveal answer

Answer

, so this can be rewritten as

$\dpi{120}=\int$\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$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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$$

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 as the sum of two fractions.

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}$2{}$ .

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}$$ .

← Didn't Know|Knew It →

0

Knew It