Int (0)^( pi)sqrt(1 cos x)dx is equal to
嚜澧ontinue
Int_(0)^( pi)sqrt(1 cos x)dx is equal to
Indeed, the integral of a function positive on $\left(0\ 2\pi\right)$-{$\pi$} can't be null. The problem is with the primitive choosen by maxima : sage: integrate(sqrt(1-cos(x)),x) -2*sqrt(2)/sqrt(sin(x)^2/(cos(x) + 1)^2 + 1) i. e. $-\frac{2 \, \sqrt{2}}{\sqrt{\frac{\sin\left(x\right)^{2}}{{\left(\cos\left(x\right) + 1\right)}^{2}} + 1}}$. This function brutally changes the sign of its
derivative for $x=\pi$. This smell of an error in manipulating an absolute value. Indeed, one can (more or less easily) show that : sage: ((cos(2*x)==cos(2*x).trig_expand().subs(cos(x)^2==1-sin(x)^2))*-1) ....: .subs(x==x/2)+1 -cos(x) + 1 == 2*sin(1/2*x)^2 Therefore, we are trying to compute $\displaystyle\int_0^{2\pi}\sqrt{2\sin(x/2)}\,dx$, which is : with
assuming(x,"real"):integrate(sqrt(2*sin(x/2)^2).simplify(),x,0,2*pi,h ....: old=True) integrate(sqrt(2)*abs(sin(1/2*x)), x, 0, 2*pi) i. e. $\int_{0}^{2 \, \pi} \sqrt{2} {\left| \sin\left(\frac{1}{2} \, x\right) \right|}\,{d x}$. since $x\in[=(0~2\pi)\Rightarrow\sin(x/2)>0$, this should be evaluated as sage: integrate(sqrt(2)*sin(x/2),x,0,2*pi,hold=True)==integrate(sqrt(2)*sin(x/2) ....:
,x,0,2*pi) integrate(sqrt(2)*sin(1/2*x), x, 0, 2*pi) == 2^(5/2) But it's not. What other possibilities do we have ? Both sympy and giac fail to find a primitive of $\sqrt{1-\cos(x)}$ or the definite integral. But fricas does find a primitive : sage: integrate(sqrt(1-cos(x)),x, algorithm="fricas") -2*(cos(x) + 1)*sqrt(-cos(x) + 1)/sin(x) i. e. $-\frac{2 \, {\left(\cos\left(x\right) +
1\right)} \sqrt{-\cos\left(x\right) + 1}}{\sin\left(x\right)}$. which can be differentiated back to the original expression : sage: integrate(sqrt(1-cos(x)),x, algorithm="fricas").diff(x).expand().trig_simp ....: lify() sqrt(-cos(x) + 1) But fricas fails to find the definite integral, which we have to find by hand : sage: foo=integrate(sqrt(1-cos(x)),x, algorithm="fricas") sage:
(foo.limit(x=2*pi,dir="-")-foo.limit(x=0,dir="+")).simplify() 4*sqrt(2) maple finds a quite different expression : sage: maple.integrate(sqrt(1-cos(x)),x).sage() 4*(cos(1/2*x)^2 - 1)*cos(1/2*x)/(sqrt(-2*cos(1/2*x)^2 + 2)*sin(1/2*x)) i. e. $\frac{4 \, {\left(\cos\left(\frac{1}{2} \, x\right)^{2} - 1\right)} \cos\left(\frac{1}{2} \, x\right)}{\sqrt{-2 \, \cos\left(\frac{1}{2} \, x\right)^{2} + 2}
\sin\left(\frac{1}{2} \, x\right)}$ whch can been redifferentiated to (something close to) the form we have seen when solving manually in order to understand maxima's error : sage: maple.integrate(sqrt(1-cos(x)),x).sage().diff(x).trig_expand().factor().tr ....: ig_simplify() sqrt(2)*sin(1/2*x)^2/sqrt(sin(1/2*x)^2) i. e. $\frac{\sqrt{2} \sin\left(\frac{1}{2} \, x\right)^{2}}
{\sqrt{\sin\left(\frac{1}{2} \, x\right)^{2}}}$. Again, since $x\in\left(0~2\pi\right)$, this amounts to $\sqrt{2\sin(x/2)}$, which we used previously. maple fails to find the definite integral, which may be recomuted by taking the limits at the bounds. mathematica gives yet another form : sage: mathematica.Integrate(sqrt(1-cos(x)),x).sage() -2*sqrt(-cos(x) + 1)*cot(1/2*x)
i. e. $-2 \, \sqrt{-\cos\left(x\right) + 1} \cot\left(\frac{1}{2} \, x\right)$. which can also be redifferentiated to the same form as maple's result : sage: mathematica.Integrate(sqrt(1-cos(x)),x).sage().subs(1-cos(x)==2*sin(x/2)^2 ....: ).diff(x).trig_simplify() sqrt(2)*sin(1/2*x)^2/sqrt(sin(1/2*x)^2) i. e. $\frac{\sqrt{2} \sin\left(\frac{1}{2} \, x\right)^{2}}{\sqrt{\sin\left(\frac{1}{2} \,
x\right)^{2}}}$. mathematica finds the definite integral : sage: mathematica.Integrate(sqrt(1-cos(x)),[x,0,2*pi]).sage() 4*sqrt(2) Not seen : the plots of the integrand and of the various solutions are also helpful to understand the problem, understand the possible errors of the various CASes and to check them against one another (hint : plot two finctions to be
compared with a slight vertical shift alowing to separate them). Numerical integration may also help... Home ? Mathematics ? Math Solution ? Integral Calculus problem ? Prev Article Next Article (Last Updated On: January 19, 2020) Problem Statement: ECE Board November 1998 Integrate the square root of (1 每 cos x) dx. Problem Answer: The integral of
the trigonometric function is -2﹟2 cos (x/2) + C. Latest Problem Solving in Integral Calculus More Questions in: Integral Calculus Online Questions and Answers in Integral Calculus Prev Article Next Article Please log in or register to answer this question. Please log in or register to add a comment. Integrals come in two varieties: indefinite and definite.
Indefinite integrals can be thought of as antiderivatives, and definite integrals give signed area or volume under a curve, surface or solid. Wolfram|Alpha can compute indefinite and definite integrals of one or more variables, and can be used to explore plots, solutions and alternate representations of a wide variety of integrals. Last updated at May 29, 2018
by Teachoo Transcript Ex 7.11, 16 By using the properties of definite integrals, evaluate the integrals : 0
log 1+ cos
Let I= 0
log 1+ cos
﹤ I= 0
1+
?
I= 0
1? cos
Adding (1) and (2) i.e. (1) + (2) I+I= 0
1+ cos
+0
1? cos
2I= 0
1+
cos
+
1? cos
2I= 0
1+ cos
1? cos
2I= 0
1? cos 2
2I= 0
2
2I= 0
2
2I=2 0
I= 0
Here,
= log sin
f ? =
?
=
sin
=
﹤ I= 0
sin
=2
0
2
sin
Let I1= 0
2
Solving I1= 0
2
﹤ I1= 0
2
2 ?
I1= 0
2
cos
Adding (2) and (3) i.e. (2) + (3) I1 + I1 = 0
2
sin
+0
2
cos
2I1 = 0
2
log sin cos
2I1 = 0
2
log 2sin cos
2
2I1 = 0
2
log 2sin cos
?
2
2I1 = 0
2
log sin 2
? 2
2I1 = 0
2
log sin 2
?0
2
log 2
Solving I2= 0
2
log sin 2
Let 2 = Differentiating both sides w.r.t. 2=
=
2 ﹤ Putting the values of t and and
changing the limits, I2 = 0
2
log sin 2
I2 = 0
log sin
2
I2 = 1 2 0
log sin
Here,
= log
2 ? = 2 ? = log
2 ?
= log sin
As,
= 2 ? ﹤ I2 = 1 2 0
log sin
= 1 2 ℅2 0
2
log sin .
=0
2
log sin .
I2= 0
2
log sin
Putting the value of I2 in equation (3), we get 2I1 = 0
2
log sin 2
?0
2
log 2
2I1 = 0
2
log sin
?log 2 0
2
1.
2I1 = I1 ? log 2
0
2
2I1?I1=?log?2
2 ?0 I1=?log?2
2
﹤ 1= ?
2 log 2 Hence, =2 I1= 2 ℅
? 2 log 2 =?
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