Question

Solve only if you know the solution and show work.

Answer 1

`\displaystyle\int(\cos x+3\sin x+7)/(\cos x+\sin x+1)\,\mathrm dx=\int\mathrm dx+2\int(\sin x+3)/(\cos x+\sin x+1)\,\mathrm dx`

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


`\sin x=\sin\left(2*\frac x2\right)=2\sin\frac x2\cos\frac x2=(2t)/(1+t^2)`

`\cos x=\cos\left(2*\frac x2\right)=\cos^2\frac x2-\sin^2\frac x2=(1-t^2)/(1+t^2)`

and


`\mathrm dt=\frac12\sec^2\frac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\frac x2\,\mathrm dt=\frac2{1+t^2}\,\mathrm dt`

Now the integral is


`\displaystyle\int\mathrm dx+2\int((2t)/(1+t^2)+3)/((1-t^2)/(1+t^2)+(2t)/(1+t^2)+1)*\frac2{1+t^2}\,\mathrm dt`

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


`\displaystyle2\int(2t+3(1+t^2))/((1-t^2+2t+1+t^2)(1+t^2))\,\mathrm dt=2\int(3t^2+2t+3)/((1+t)(1+t^2))\,\mathrm dt`

Decompose the integrand into partial fractions:


`(3t^2+2t+3)/((1+t)(1+t^2))=\frac2{1+t}+(1+t)/(1+t^2)`

so you have


`\displaystyle2\int(3t^2+2t+3)/((1+t)(1+t^2))\,\mathrm dt=4\int(\mathrm dt)/(1+t)+2\int(\mathrm dt)/(1+t^2)+\int(2t)/(1+t^2)\,\mathrm dt`

which are all standard integrals. You end up with


`\displaystyle\int\mathrm dx+4\int(\mathrm dt)/(1+t)+2\int(\mathrm dt)/(1+t^2)+\int(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\frac x2\right|+2\arctan\left(\arctan\frac x2\right)+\ln\left(1+\tan^2\frac x2\right)+C`

`=2x+4\ln\left|1+\tan\frac x2\right|+\ln\left(\sec^2\frac 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\frac x2\right|` to get


`2x+5\ln\left|1+\tan\frac x2\right|+\ln\left(\sec^2\frac x2\right)-\ln\left|1+\tan\frac x2\right|+C`

`2x+5\ln\left|1+\tan\frac x2\right|+\ln\left|(\sec^2\frac x2)/(1+\tan\frac x2)\right|+C`

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


`(\sec^2\frac x2)/(1+\tan\frac x2)=\sin x+\cos x+1`

You have


`(\sec^2\frac x2)/(1+\tan\frac x2)=\frac1{\cos^2\frac x2+\sin\frac x2\cos\frac x2}`

`(\sec^2\frac x2)/(1+\tan\frac x2)=\frac1{\frac{1+\cos x}2+\frac{\sin x}2}`

`(\sec^2\frac x2)/(1+\tan\frac x2)=\frac2{\cos x+\sin x+1}`

So in the corresponding term of the antiderivative, you get


`\ln\left|(\sec^2\frac x2)/(1+\tan\frac x2)\right|=\ln\left|\frac2{\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(\cos x+3\sin x+7)/(\cos x+\sin x+1)\,\mathrm dx=2x+5\ln\left|1+\tan\frac x2\right|-\ln|\cos x+\sin x+1|+C`