The correct option is A I = 2 tan√(x)+c I=∫ sec ^ 2 √( x ) #160; /√( x ) #160; #160; dx Let u=√( x )⇒ du= 1 / 2√( x ) #160; dx Subs ...

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Integration of Trigonometric Functions

Integrate: ...

Question

Integrate: $\int \frac{{sec}^{2} \sqrt{x}}{\sqrt{x}} d x$

I = 2 tan \sqrt{x} + c

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I = 2 cot \sqrt{x} + c

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I = 3 tan \sqrt{x} + c

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I = 2^{2} tan \sqrt{x} + c

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Solution

The correct option is A $I = 2 tan \sqrt{x} + c$
$I = \int \frac{{s e c}^{2} \sqrt{x}}{\sqrt{x}} d x L e t u = \sqrt{x} \Rightarrow d u = \frac{1}{2 \sqrt{x}} d x$
Substituting in I,

$I = 2 \int {s e c}^{2} u d u = 2 t a n u + c [∵ \frac{d (t a n u)}{d u} {s e c}^{2} u]$

$\Rightarrow I = 2 t a n \sqrt{x} + c$

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