Question

can you solve this integral​

Answer 1

Answer:

$\green{ \boxed{\displaystyle \int \bf \: \frac{sin \sqrt{x} }{ \sqrt{x} } = - 2cos \: \sqrt{x} + c \: }}$

Step-by-step explanation:

$\displaystyle \int \bf \: \frac{sin \sqrt{x} }{ \pink{\sqrt{x} } } \: \pink{ dx }\\ \\ \bf \: substitute \: \: u = \sqrt{x } \\ \bf \implies \: \frac{du}{dx} = \frac{1}{ 2\sqrt{x} } \\ \bf \implies \: 2 \: du = \pink{ \frac{dx}{ \sqrt{x} } } \\ \\ = \displaystyle 2\int \bf \: \: sin \: u \: du \\ \\ = \bf - 2 \: cos \: u + c\: \: \: \: \red{\{ \because \: \displaystyle \int \bf \:sin \: x \: dx = \: - cos \: x + c \: \}} \\ \\ \bf = - 2cos \sqrt{x} + c$

Extra formula :

$\odot\: \: \: \displaystyle \bf\int \dfrac{1}{x} \: dx =log |x |+ C \\ \\ \odot \: \: \: \displaystyle \bf\int {a}^{x} \: dx = \frac{ {a}^{x} }{log \: a} + C \\ \\ \odot \: \: \: \displaystyle \bf\int sinx \: dx = - cosx + C \\ \\ \odot\: \: \: \displaystyle \bf\int cosx \: dx =sinx + C \\ \\ \odot \: \: \: \displaystyle \bf\int sec {}^{2} x \: dx =tanc + C \\ \\ \odot \: \: \: \displaystyle \bf\int {cosec}^{2} x \: dx = - cotx + C \\ \\ \odot \: \: \: \displaystyle \bf\int secx.tanx \: dx =secx + C \\ \\ \odot \: \: \: \displaystyle \bf\int cosecx.cotx \: dx = - cosecx + C \:$