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Murljashka [212]
3 years ago
7

Solve only if you know the solution and show work.

Mathematics
1 answer:
SashulF [63]3 years ago
5 0
\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx

For the remaining integral, let t=\tan\dfrac x2. Then

\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}
\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}

and

\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt

Now the integral is

\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt

The first integral is trivial, so we'll focus on the latter one. You have

\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt

Decompose the integrand into partial fractions:

\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}

so you have

\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt

which are all standard integrals. You end up with

\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt
=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C
=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C
=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C

To try to get the terms to match up with the available answers, let's add and subtract \ln\left|1+\tan\dfrac x2\right| to get

2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C
2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C

which suggests A may be the answer. To make sure this is the case, show that

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1

You have

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}

So in the corresponding term of the antiderivative, you get

\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|
=\ln2-\ln|\cos x+\sin x+1|

The \ln2 term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C
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Answer:

Step-by-step explanation:

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3 0
2 years ago
Explain how to find an estimate for the quotient 3.4 ÷ 6
vovikov84 [41]
To estimate this we can do it like this:-
<span>
Way #1

3.4 = 3
6 = 6 

3 </span>÷ 6 = <span>0.5

So, to estimate 3.4 </span>÷ 6 we round 3.4 to the nearest whole number and den divide. Den d answer is our estimated answer!

3.4 ÷ 6 = 0.5

Hope I helped ya!! 
3 0
3 years ago
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NEED HELP ASAP<br> Solve the equation or inequality for the unknown number. Show your work.
amid [387]

Answer:

5

Step-by-step explanation:

3(14+x) = 57

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6 0
3 years ago
Please help me with this thanks
Kryger [21]

Step-by-step explanation:

B = area of the base

= 9×4 = 36 square inches

P= perimeter of the base

= 2(9+4)

=2×13= 26 inches

H =distance between the bases

= 16 inches

SA= 2B+PH

= 2(36)+ 26× 16

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8 0
3 years ago
F(x)=2x+3/4x+5<br>find f(-9)<br>​
mylen [45]

\implies {\blue {\boxed {\boxed {\purple {\sf { f(-9)= 0.48}}}}}}

\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}

f(x) =  \frac{2x + 3}{4x + 5} \\

For f(-9), put "-9" for every value of "x".

↬f( - 9) =  \frac{2( - 9) + 3}{4( - 9) + 5}\\

↬ f(-9) =  \frac{ - 18 + 3}{ - 36 + 5} \\

↬ f(-9) =  \frac{ - 15}{ - 31}\\

↬ f(-9)=  \frac{15}{31}\\

↬f(-9)=  0.48\\

\bold{ \green{ \star{ \red{Mystique35}}}}⋆

6 0
3 years ago
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