-cos-x-x d x-A- 2 sin-x-cB- sin-x-cC- -2 cos-x-cD- -cos-x-c

The correct option is A Put √(x)=t ⇒ x=t^2 dx=2tdt G.I. ∫cos t/t 2tdt = 2sin t +C = 2 sin √(x) + C ...

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Algebra of Derivatives

∫cos√x/√x d x...

Question

$\int \frac{c o s \sqrt{x}}{\sqrt{x}} d x =$

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$\int \frac{c o s \sqrt{x}}{\sqrt{x}} d x =$

c o s^{- 1} x = 2 s i n^{- 1} \sqrt{\frac{1 - x}{2}} = 2 c o s^{- 1} \sqrt{\frac{1 + x}{2}}

s i n^{- 1} (- x) = - s i n^{- 1} x, c o s^{- 1} (- x) = π - c o s^{- 1} x (- 1 \leq x \leq 1)

\int \sqrt{\frac{x}{1 - x}} d x

\sin^{- 1} \sqrt{x} + C

\sin^{- 1} \{\sqrt{x} - \sqrt{x (1 - x)}\} + C

\sin^{- 1} \{\sqrt{x (1 - x)}\} + C

\sin^{- 1} \sqrt{x} - \sqrt{x (1 - x)} + C

\int \frac{s e n x d x}{\sqrt{s i n (2 x + A) + s i n A}} = \int \frac{s e c x d x}{\sqrt{2 s i n x c o s x + (2 c o s^{2} x - 1) s i n A + s i n A}} = \int \frac{s e c x d x}{\sqrt{2 c o s x s i n (x + A)}}

\int \frac{s i n^{- 1} \sqrt{x} - c o s^{- 1} \sqrt{x}}{s i n^{- 1} \sqrt{x} + c o s^{- 1} \sqrt{x}} d x = \frac{A}{748 π} [- (s i n^{- 1} \sqrt{x}) (1 - 2 x) + \sqrt{x} \sqrt{1 - x}] - x + C

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