The correct option is C 2[sin √(x) √(x) cos √(x)]+cPut √(x) = t ⇒1/2 √(x) dx = dt ⇒ dx = 2 t dt, then ∫ sin √(x) dx = 2 ∫ t sin t dt = 2( t cos t+sin t)+c =2 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫sin√x d x=

Question

$\int s i n \sqrt{x} d x =$

2 [s i n \sqrt{x} - c o s \sqrt{x}] + c

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2 [s i n \sqrt{x} - \sqrt{x} c o s \sqrt{x}] + c

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2 [s i n \sqrt{x} + c o s \sqrt{x}] + c

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2 [s i n \sqrt{x} + \sqrt{x} c o s \sqrt{x}] + c

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Solution

The correct option is B $2 [s i n \sqrt{x} - \sqrt{x} c o s \sqrt{x}] + c$
Put $\sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 t d t$ , then
$\int s i n \sqrt{x} d x = 2 \int t s i n t d t = 2 (- t c o s t + s i n t) + c$
$= 2 (s i n \sqrt{x} - \sqrt{x} c o s \sqrt{x}) + c$

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∫sin√x dx=

The correct option is B 2[sin√x−√xcos√x]+cPut √x=t⇒12√x dx=dt⇒dx=2 t dt, then ∫sin√x dx=2∫t sin t dt=2(−t cos t+sin t)+c =2(sin√x−√x cos√x)+c

$\int s i n \sqrt{x} d x =$