Home > Calculus > Integrals of Trigonometric Functions

How do you integrate #int cosx/sqrt(1-2sinx) #?

Answer 1

Charles Arocha

#-sqrt(1-2sinx)+C#

We have:

#intcosx/sqrt(1-2sinx)dx#

Let: #u=1-2sinx#. This implies that #du=-2cosxdx#.

#intcosx/sqrt(1-2sinx)dx=-1/2int(-2cosx)/sqrt(1-2sinx)dx#

#=-1/2int(du)/sqrtu=-1/2intu^(-1/2)du=-1/2(u^(-1/2+1)/(-1/2+1))+C#

#=-u^(1/2)+C=-sqrt(1-2sinx)+C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emma Aldridge

To integrate ( \int \frac{\cos x}{\sqrt{1 - 2 \sin x}} ), we can use a trigonometric substitution.

Let ( u = 1 - 2 \sin x ), then ( du = -2 \cos x dx ).

Substitute ( du/(-2) = \cos x dx ) and ( 1 - u = 2 \sin x ) into the integral:

[ \int \frac{\frac{du}{-2}}{\sqrt{u}} ]

This simplifies to:

[ -\frac{1}{2} \int \frac{1}{\sqrt{u}} , du ]

Now integrate:

[ -\frac{1}{2} \int u^{-\frac{1}{2}} , du ]

[ -\frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C ]

[ -\sqrt{u} + C ]

Now, substitute back ( u = 1 - 2 \sin x ):

[ -\sqrt{1 - 2 \sin x} + C ]

So, ( \int \frac{\cos x}{\sqrt{1 - 2 \sin x}} , dx = -\sqrt{1 - 2 \sin x} + C ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.