Question

Solve integral

Answer 1

Where is the question

Answer 2

\[\int\limits dx/(cosx+2sinx+3)\]

Answer 3

\[-sinx+cosx/2+3x\]

Answer 4

I still don't know what you mean

Answer 5

\[\int\limits\frac{dx}{cosx+2sinx+3} \]
weierstrass substitution might work..

Answer 6

i just know that the answer is arctan( 1+ tan(x/2) ) + c
but i don't know the solution

Answer 7

@soati : what do you think about it?

Answer 8

I think I got it. Give me a minute to write it down. I like your integrals, theyre difficult!

Answer 9

let t=tg(x/2) => x=2arctgt and dx=2dt/(1+ t^2)
by sinx=2t/(1+t^2), cosx=(1-t^2)/(1+t^2)
then you can solve it

Answer 10

Let \[u = \tan ({x \over 2})\]
\[dx = 2 \cos^2({x \over 2}) du\]
\[du = {1\over2}{1 \over \cos^2({x \over 2})}dx\]
\[\int\limits {dx \over cosx+2sinx+3} = \int\limits {2 \cos^2({x \over 2}) \over cosx+2sinx+3}du\]

Use the trig. rules:
\[\cos(u \pm v) = \cos(u)\cos(v) \mp \sin(u)\sin(v)\]
\[\sin(u \pm v) = \sin(u)\cos(v) \pm \cos(u)\sin(v)\]

In our case:
\[\cos(x) = \cos({x \over 2} + {x \over 2}) = \cos^2({x \over 2}) - \sin^2({x \over 2}) = 2\cos^2({x \over 2}) -1\]
\[\sin(x) = \sin({x \over 2} + {x \over 2}) =2 \sin({x \over 2}) \cos({x \over 2})\]

\[\int\limits\limits {2 \cos^2({x \over 2}) \over cosx+2sinx+3}du = \int\limits\limits {2 \cos^2({x \over 2}) \over 2\cos^2({x \over 2})+2*2\sin({x \over 2})\cos({x \over 2})+3 - 1}du \]
\[\int\limits\limits {1\over 1+2{\sin({x \over 2}) \over \cos({x \over 2}) }+{1\over \cos^2({x \over 2})}}du \]

Knowing that..
\[{1 \over \cos(u)} = \sec(x)\]
\[1 + \tan^2(u) = \sec^2(u)\]

\[\int\limits\limits\limits {1\over 1+2{\sin({x \over 2}) \over \cos({x \over 2}) }+{1\over \cos^2({x \over 2})}}du = \int\limits\limits\limits {1\over 1+2\tan({x \over 2}) +{\sec^2({x \over 2})}}du = \int\limits\limits\limits {1\over 1+2\tan({x \over 2}) +\tan^2({x \over 2}) + 1}du \]

Finally replace tan(x/2) by u!
\[\int\limits\limits\limits\limits {1\over 1+2u +u^2 + 1}du =\int\limits\limits\limits\limits {1\over (u+1)^2 + 1}du\]

We know that
\[\int\limits\limits\limits\limits {1\over u^2 +1}du = \arctan(u)\]

So we get
\[\int\limits\limits\limits\limits {1\over (u+1)^2 + 1}du = arctan( u+1 )\]

Replace u by tan(x/2)...
\[\int\limits\limits {dx \over cosx+2sinx+3} = \arctan(1+\tan({x \over 2}) ) + C\]

Answer 11

thank @soati and @RATANAK :)