A = 3. Hope this helps! (Was trying to show the work but I can't take pictures at the moment.)
Answer:
+
*LN(|
|) +C
Step-by-step explanation:
we will have to do a trig sub for this
use x=a*tanθ for sqrt(x^2 +a^2) where a=2
x=2tanθ, dx= 2 sec^2 (θ) dθ
this turns
into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ
the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1
then it simplifies into integral(4*sec^3 (θ)) dθ
you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C
then you will need to rework your functions of θ back into functions of x
tanθ will resolve back into
(see substitutions) while secθ will resolve into
sec(θ)=
is from its ratio identity of hyp/adj where the hyp. is
and adj is 2 (see tan(θ) ratio)
after resolving back into functions of x, substitute ratios for trig functions:
=
+
*LN(|
|) +C
Answer:
Step-by-step explanation:
\[2~ sin^2 x+3sin x+1=0\]
\[2sin^2x+2sin x+sin x+1=0\]
2sinx(sin x+1)+1(sin x+1)=0
(sin x+1)(2 sin x+1)=0
either sin x+1=0
sin x=-1=sin 3π/2=sin (2nπ+3π/2)
x=2nπ+3π/2,where n is an integer.
or 2sin x+1=0
sin x=-1/2=-sin π/6=sin (π+π/6),sin (2π-π/6)=sin (2nπ+7π/6),sin (2nπ+11π/6)
x=2nπ+7π/6,2nπ+11π/6,
where n is an integer.
Answer:
sure
Step-by-step explanation: