-1-2 x-tan2 xdx-log-sin x-fx-C- then which of the following is-are correct-

∫cos x√(1+sin^2 x)dx⇒sin x=t ⇒cos x dx =dt =∫dt√(1+t^2)=log | t+√(1+t^2) |+C =log | sin x+√(1+sin^2 x) |+C ⇒ f(x)=1+sin^2x So, f_min=1, f_max=2 and f(0)=1 ...

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Byju's Answer

Standard XII

Physics

The Problem of Areas

∫1√2 x+tan2 x...

Question

$\int \frac{1}{\sqrt{{sec}^{2} x + {tan}^{2} x}} d x = log | sin x + \sqrt{f (x)} | + C$ , then which of the following is/are correct?

f_{m i n} = 0

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f_{m i n} = 1

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f_{m a x} = 2

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f (0) = 1

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Solution

The correct option is D $f (0) = 1$
$\int \frac{cos x}{\sqrt{1 + {sin}^{2} x}} d x \Rightarrow sin x = t \Rightarrow cos x d x = d t = \int \frac{d t}{\sqrt{1 + t^{2}}} = log ∣ ∣ t + \sqrt{1 + t^{2}} ∣ ∣ + C = log ∣ ∣ sin x + \sqrt{1 + {sin}^{2} x} ∣ ∣ + C$
$\Rightarrow f (x) = 1 + {sin}^{2} x$
So, $f_{m i n} = 1, f_{m a x} = 2$ and $f (0) = 1$

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Similar questions

\int \frac{1}{\sqrt{{sec}^{2} x + {tan}^{2} x}} d x = log | f (x) + \sqrt{g (x)} | + C

, then which of the following is/are correct?

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___

Q. (A) :

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Q. If

\int \frac{cos x}{sin (x - \frac{π}{6}) sin (x + \frac{π}{6})} d x = log ∣ ∣ ∣ \frac{2 f (x) - 1}{2 f (x) + 1} ∣ ∣ ∣ + C,

then which of the following is/are true?

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∫1√sec2x+tan2xdx=log|sinx+√f(x)|+C, then which of the following is/are correct?

The correct option is D f(0)=1∫cosx√1+sin2xdx⇒sinx=t⇒cosx dx=dt=∫dt√1+t2=log∣∣t+√1+t2∣∣+C=log∣∣sinx+√1+sin2x∣∣+C ⇒f(x)=1+sin2x So, fmin=1,fmax=2 and f(0)=1

$\int \frac{1}{\sqrt{{sec}^{2} x + {tan}^{2} x}} d x = log | sin x + \sqrt{f (x)} | + C$ , then which of the following is/are correct?