Question

Find : $\int \sqrt{1+cosx}\:dx$​

Answer 1

Heya!!

Before doing its integration let's make this function in simpler form

√{1 + Cos x } = √{ 2 Cos² x/2 } =

√2 × Cos x/2

=>

Integration of √{1 + Cos x } dx = integration of √2 × Cos x/2 dx

=>

2√2 × Sin x/2 + C

Where c is called constant of integration.

Answer 2

given★

$\int \sqrt{1+cosx}\:dx$

solution★

• $\int \sqrt{1+cosx}\:dx$

we know that

• $\cos(2x) = 2 \cos {}^{2} (x) - 1 \\ \\ 1 + \cos(2x) = 2 \cos {}^{2} (x) \\ \\ angle \: half \\ \\ 1 + \cos(x) = 2 \cos {}^{2} ( \frac{x}{2} )$
• $\int \sqrt{2cos^2x/2}\:dx$

• $\int{cosx/2√2}\:dx$

• we know that
• $\int{cosx}\:dx$=sinx+c

============================

• 2√2 sinx/2+c

• where c constant

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I hope it helps you