What is $f (x) = \int x {cot x}^{2} d x$ $f(x) = int xcotx^2 dx$ if $f (\frac{5 π}{4}) = 0$ $f((5pi)/4) = 0$ ?

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What is $f (x) = \int x {cot x}^{2} d x$ $f(x) = int xcotx^2 dx$ if $f (\frac{5 π}{4}) = 0$ $f((5pi)/4) = 0$ ?

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Answer 1 · 2018-05-05T07:32:54

I have tried , but it is complicated for $x^{2} = {(\frac{5 π}{4})}^{2} .$ $x^2=((5pi)/4)^2.$

$f (x) = \frac{1}{2} ln ∣ {sin x}^{2} ∣ ∣ + c, w h e r e, c \approx 0.6623$ $f(x)=1/2ln|sinx^2|+c,where, c~~0.6623$

Explanation:

Here,

$f (x) = I = \int x {cot x}^{2} d x = \int {cot x}^{2} \cdot x d x$ $f(x)=I=intxcotx^2dx=intcotx^2*xdx$

Let, $x^{2} = u \Rightarrow 2 x d x = d u \Rightarrow x d x = \frac{1}{2} d u$ $x^2=u=>2xdx=du=>xdx=1/2du$

So,

$I = \int cot u \cdot \frac{1}{2} d u$ $I=intcotu*1/2du$

$= \frac{1}{2} ln | sin u | + c, w h e r e, u = x^{2}$ $=1/2ln|sinu|+c,where,u=x^2$

$=>f(x)=1/2ln|sinx^2|+c...to(A)$

Given that,

$f((5pi)/4)=0$

$=>1/2ln|sin((5pi)/4)^2|+c=0$

$=>2c=-ln|sin((25pi^2)/16)|$

$=>2c=ln|(1/(sin((25pi^2)/16)))|$

$=>c=1/2ln|(1/(sin((25pi^2)/16)))|...to(1)$

Now,

$(25xxpi^2)/16 ~~15.42....$

$sin((25pi^2)/16)~~0.2659... in[-1,1]$

$1/(sin((25pi^2)/16))~~3.7606...$

$ln|1/(sin((25pi^2)/16))| ~~1.3245...$

$1/2*ln|1/(sin((25pi^2)/16))| ~~0.6623$

Hence, from $(1)$

$c ~~$ 0.6623

Thus,

$f(x)=1/2ln|sinx^2|+c,where, c~~0.6623$

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What is $f (x) = \int x {cot x}^{2} d x$ $f(x) = int xcotx^2 dx$ if $f (\frac{5 π}{4}) = 0$ $f((5pi)/4) = 0$ ?

1 Answer

Explanation:

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Explanation:

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What is $f (x) = \int x {cot x}^{2} d x$ $f(x) = int xcotx^2 dx$ if $f (\frac{5 π}{4}) = 0$ $f((5pi)/4) = 0$ ?