Couple of integrals: int [1/(1+cosx)] dx, etc.

[MarkSA]

Hello,

I have a few "problem" integrals that i'm having difficulty solving:

1) integral: [1/(1+cosx)]dx
1/(1+cosx) is the same as writing: (1 + cosx)^-1
However, wouldn't the antiderivative of that need to include (1 + cosx)^0, which must be equal to 1? Is this one a typo on my paper or is there something i'm missing?

2) integral: [3x*sqrt(1-2x^2)]dx
This is the same as writing: 3x*sqrt(1-2x^2)^-1. But I can't wrap my mind around how to find the antiderivative with this being a product, since I can't multiply the 3x in...

3) integral: [(5x^2+1)(5x^3+3x-8)^6]dx
Yikes! that to the 6th power is not something I want to multiply out in order to find the antiderivative of. Is there an easier way to do it?

 

[soroban]

Re: Couple of integrals

Hello, MarkSA!

\(\displaystyle 1)\;\;\L\int \frac{1}{1\,+\,\cos x}\,dx\)

Double-angle identity: \(\displaystyle \:\cos^2\frac{x}{2} \:=\:\frac{1\,+\,\cos x}{2}\;\;\Rightarrow\;\;1\,+\,\cos x \:=\:2\cdot\cos^2\left(\frac{x}{2}\right)\)

So we have: \(\displaystyle \L\:\int\frac{1}{1\,+\,\cos x}\,dx \;=\;\int\frac{1}{2\cdot\cos^2\left(\frac{x}{2}\right)}\,dx \;=\;\frac{1}{2}\int\sec^2\left(\frac{x}{2}\right)\,dx\)

Got it?

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Another approach . . .

Multiply top and bottom by \(\displaystyle (1\,-\,\cos x)\)

\(\displaystyle \L\frac{1\,-\,\cos x}{1\,-\,\cos x}\,\cdot\,\frac{1}{1\,+\,\cos x} \;=\;\frac{1\,-\,\cos x}{1\,-\,\cos^2x}\;=\;\frac{1\,-\,cos x}{\sin^2x}\)

. . \(\displaystyle \L=\;\frac{1}{\sin^2x}\,-\,\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} \;=\;\csc^2x\,-\,\csc x\cdot\cot x\)

And you can intergrate: \(\displaystyle \L\:\int \left(\csc^2x\,-\,\csc x\cdot\cot x\right)\,dx\)

\(\displaystyle 2)\L\;\;\int 3x\sqrt{1\,-\,2x^2}\,]dx\)

We have: \(\displaystyle \L\:\int 3x(1\,-\,2x^2)^{\frac{1}{2}}\,dx\)

Let: \(\displaystyle u \:=\:1\,-\,2x^2\)

\(\displaystyle 3)\;\;\L\int (5x^2\,+\,1)(5x^3\,+\,3x\,-\,8)^6\,dx\)

Let: \(\displaystyle \,u \:=\:5x^3\,+\,3x\,-\,8\)

 

[MarkSA]

thank you