If -sec2 x -2010-sin2010xdx-Px-sin2010x-C- then value of P -3 is

The correct option is C √(3)∫sec^2 x 2010/sin^2010xdx =∫ sec^2 x(sin x)^ 2010 2010 ∫1/(sin x)^2010dx=I_1 I_2 Applying by parts on I_1, we get I_1=tan x/(sin ...

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Standard XII

Mathematics

Integration by Parts

If ∫sec2 x -2...

Question

If $\int \frac{s e c^{2} x - 2010}{s i n^{2010} x} d x = \frac{P (x)}{s i n^{2010} x} + C,$ then value of $P (\frac{π}{3})$ is

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\frac{1}{\sqrt{3}}

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\sqrt{3}

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Solution

The correct option is C $\sqrt{3}$
$\int \frac{s e c^{2} x - 2010}{s i n^{2010} x} d x = \int s e c^{2} x (s i n x)^{- 2010} - 2010 \int \frac{1}{(s i n x)^{2010}} d x = I_{1} - I_{2}$
Applying by parts on $I_{1}$ , we get
$I_{1} = \frac{t a n x}{(s i n x)^{2010}} + 2010 \int \frac{t a n x c o s x}{(s i n x)^{2011}} d x = \frac{t a n x}{(s i n x)^{2010}} + 2010 \int \frac{d x}{(s i n x)^{2010}} \Rightarrow I = I_{1} - I_{2} = \frac{t a n x}{(s i n x)^{2010}} = \frac{P (x)}{(s i n x)^{2010}} P (\frac{π}{3}) = t a n \frac{π}{3} = \sqrt{3}$

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If ∫sec2x−2010sin2010xdx=P(x)sin2010x+C, then value of P(π3) is

The correct option is C √3∫sec2x−2010sin2010xdx=∫sec2x(sin x)−2010−2010∫1(sin x)2010dx=I1−I2 Applying by parts on I1, we get I1=tanx(sinx)2010+2010∫tanx cosx(sinx)2011dx=tanx(sinx)2010+2010∫dx(sinx)2010⇒I=I1−I2=tanx(sinx)2010=P(x)(sinx)2010P(π3)=tanπ3=√3

If $\int \frac{s e c^{2} x - 2010}{s i n^{2010} x} d x = \frac{P (x)}{s i n^{2010} x} + C,$ then value of $P (\frac{π}{3})$ is