If - x -log sin x - x - x - n - ne I and -133 log sin x - x dx - k -1 then the value of k isA- 27B- 18C- 9D- None of these

The correct option is A 27∫_1^3 3 log(sin x^3)/xdx=∫_1^27log(sinz)/zdz [ putting x^3=z ⇒ 3dx=dz/x^2 ] =∫_1^27Φ (z)dz=[Φ (z)]_1^27=Φ (27) Φ (1) ∴ k=27 ...

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Byju's Answer

Standard XII

Mathematics

Property 1

If Φ' x =log ...

Question

If $Φ^{'} (x) = \frac{l o g (s i n x)}{x}, x \neq n π, n ϵ I a n d \int_{1}^{3} \frac{3 l o g (s i n x)}{x} d x = Φ (k) - Φ (1)$ then the value of k is

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None of these

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Solution

The correct option is A 27
$\int_{1}^{3} \frac{3 l o g (s i n x^{3})}{x} d x = \int_{1}^{27} \frac{l o g (s i n z)}{z} d z$
$[p u t t i n g x^{3} = z \Rightarrow 3 d x = \frac{d z}{x^{2}}]$
$= \int_{1}^{27} Φ (z) d z = [Φ (z)]_{1}^{27} = Φ (27) - Φ (1)$
$∴ k = 27$

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If Φ′(x)=log(sin x)x,x≠n π, nϵ I and ∫313log(sin x)xdx=Φ(k)−Φ(1) then the value of k is

The correct option is A 27∫313 log(sin x3)xdx=∫271log(sinz)zdz [putting x3=z⇒3dx=dzx2] =∫271Φ(z)dz=[Φ(z)]271=Φ(27)−Φ(1) ∴k=27

If $Φ^{'} (x) = \frac{l o g (s i n x)}{x}, x \neq n π, n ϵ I a n d \int_{1}^{3} \frac{3 l o g (s i n x)}{x} d x = Φ (k) - Φ (1)$ then the value of k is

The correct option is A 27
$\int_{1}^{3} \frac{3 l o g (s i n x^{3})}{x} d x = \int_{1}^{27} \frac{l o g (s i n z)}{z} d z$
$[p u t t i n g x^{3} = z \Rightarrow 3 d x = \frac{d z}{x^{2}}]$
$= \int_{1}^{27} Φ (z) d z = [Φ (z)]_{1}^{27} = Φ (27) - Φ (1)$
$∴ k = 27$