sin-x-cos-x-x-sin-2-x-dx-

Question Number 137509 by liberty last updated on 03/Apr/21

\int \frac{\sin (\sqrt{x}) + \cos (\sqrt{x})}{\sqrt{x} \sin (2 \sqrt{x})} d x

Commented by liberty last updated on 04/Apr/21

Answered by Ñï= last updated on 03/Apr/21

\int \frac{\sin \sqrt{x} + \cos \sqrt{x}}{\sqrt{x} \sin (2 \sqrt{x})} d x = 2 \int \frac{\sin u + \cos u}{\sin 2 u} d u \dots \dots . . u = \sqrt{x}

= \int (\frac{1}{\cos u} + \frac{1}{\sin u}) d

= l n ∣ s e c u + \tan u ∣ + l n ∣ c s c u - \cot u ∣ + C

= l n ∣ \sec \sqrt{x} + \tan \sqrt{x} ∣ + l n ∣ c s c \sqrt{x} - \cot \sqrt{x} ∣ + C

Answered by mathmax by abdo last updated on 03/Apr/21

I = \int \frac{\sin (\sqrt{x}) + \cos (\sqrt{x})}{\sqrt{x} \sin (2 \sqrt{x})} dx \Rightarrow I =_{\sqrt{x} = t} \int \frac{sint + cost}{tsin (2 t)} (2 t) dt

= 2 \int \frac{sint + cost}{2 sint . cost} dt = \int \frac{dt}{cost} + \int \frac{dt}{sint}

changement \tan (\frac{t}{2}) = y give \int \frac{dt}{cost} = \int \frac{2 dy}{(1 + y^{2}) \frac{1 - y^{2}}{1 + y^{2}}}

= 2 \int \frac{dy}{(1 - y) (1 + y)} = \int (\frac{1}{1 - y} + \frac{1}{1 + y}) dy = \log ∣ \frac{1 + y}{1 - y} ∣ + c_{1}

= \log ∣ \frac{1 + \tan (\frac{\sqrt{x}}{2})}{1 - \tan (\frac{\sqrt{x}}{2})} ∣ + c_{1}

\int \frac{dt}{sint} = \int \frac{2 dy}{(1 + y^{2}) . \frac{2 y}{1 + y^{2}}} = \int \frac{dy}{y} = \log ∣ y ∣ + c_{2}

= \log ∣ \tan (\frac{\sqrt{x}}{2}) ∣ + c_{2} \Rightarrow

I = \log ∣ \frac{1 + \tan (\frac{\sqrt{x}}{2})}{1 - \tan (\frac{\sqrt{x}}{2})} ∣ + \log ∣ \tan (\frac{\sqrt{x}}{2}) ∣ + C