What is #int \ (sinx+cosx)/sqrt(1+sin2x) \ dx #?

Question

Evelyn Agard · Accepted Answer

#((sinx+cosx)*dx)/sqrt(1+sin2x)=x+c# #((sinx+cosx)*dx)/sqrt(1+sin2x)# =#((sinx+cosx)*dx)/sqrt((sinx+cosx)^2)# =#((sinx+cosx)*dx)/(sinx+cosx)# =#int dx# =#x+c#

William Abdalla · Answer

# int \ (sinx+cosx)/sqrt(1+sin2x) \ dx = arctan tanx + C #

In addition to the existing solution:

We seek:

# I = int \ (sinx+cosx)/sqrt(1+sin2x) \ dx #

Using the trigonometric Identity:

# sin^2A+ cos^2A -= 1#

We can manipulate the integral as follows:

# I = int \ (sinx+cosx)/sqrt(1+2sinxcosx) \ dx #
# \ \ = int \ (sinx+cosx)/sqrt(sin^2x+ cos^2x+2sinxcosx) \ dx #
# \ \ = int \ (sinx+cosx)/(sqrt((sinx+cosx)^2)) \ dx # ..... [A]
We may be tempted to continue as follows:

# I = int \ (sinx+cosx)/(sqrt((sinx+cosx)^2)) \ dx #
# \ \ = int \ (sinx+cosx)/(sinx+cosx) \ dx #
# \ \ = int \ 1 \ dx #
# \ \ = x+c # ..... [B]
However, There now two issues that require further analysis:

(1) There are discontinuities in the integrand when:

# sinx+cosx = 0 => tanx=-1 => x=npi-pi/4 #
e.g at #x=(-5pi)/4, -pi/4, (3pi)/4, (7pi)/4 #

(2) Any linear combination of periodic functions is itself periodic, further, any linear combination of sinusoidal functions is also sinusoidal. The introduction of the square in the denominator causes it be positive #AA x in RR#, thus:

if #{ (sinx+cosx<0, =>, "Integrand " < 0), (sinx+cosx=0, =>, "Integrand is undefined"), (sinx+cosx>0, =>, "Integrand " > 0) :} #

We can graph the function as follows, and we discover it represents a "saw tooth" with period #pi# and amplitude #a#

Consequently, we expect the solution to the integral to be periodic, and this is not reflected in the naive solution [B]

Instead, we evaluate the integral as follows:

# I = int \ (sinx+cosx)/sqrt(1+2sinxcosx) \ dx #

Perform a substitution #u=tan x# so that:

# I = int \ 1/(1+u^2) \ du #
# \ \ = arctan u + C #
# \ \ = arctan tanx + C #
# \ \ != x + C \ \ AA x in RR#

And we can graph the the integral result (in red) alongside the original periodic function:

And we see that we get a periodic sawtooth function appropriate for the area under a square wave function.

Scarlett Allison · Answer

undefined To find $ \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx $, we can start by simplifying the integrand. We'll use the identity $ \sin 2x = 2\sin x \cos x $ to rewrite the denominator.

$$ \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx $$

$$ = \int \frac{\sin x + \cos x}{\sqrt{1 + 2\sin x \cos x}} \, dx $$

Next, we'll multiply the numerator and denominator by $ \sqrt{2} $ to simplify further.

$$ = \int \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2(1 + 2\sin x \cos x)}} \, dx $$

$$ = \int \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 + 4\sin x \cos x}} \, dx $$

Now, we'll use a substitution. Let $ u = \sin x + \cos x $, then $ du = (\cos x - \sin x) \, dx $. This suggests that we need a factor of $ \sqrt{2} $ in the numerator. So, we rewrite the integral as:

$$ = \frac{1}{\sqrt{2}} \int \frac{du}{\sqrt{2 + 4u^2}} $$

$$ = \frac{1}{\sqrt{2}} \int \frac{du}{\sqrt{1 + 2u^2}} $$

Now, we can use the substitution $ \sqrt{2}u = \sqrt{2} \tan \theta $, then $ \sqrt{2}du = \sqrt{2} \sec^2 \theta d\theta $, to simplify the integral.

$$ = \frac{1}{\sqrt{2}} \int \frac{\sqrt{2}\sec^2 \theta d\theta}{\sqrt{1 + 2(\tan \theta)^2}} $$

$$ = \frac{1}{\sqrt{2}} \int \frac{\sec^2 \theta d\theta}{\sqrt{\sec^2 \theta}} $$

$$ = \frac{1}{\sqrt{2}} \int \sec \theta d\theta $$

$$ = \frac{1}{\sqrt{2}} \ln|\sec \theta + \tan \theta| + C $$

Finally, we substitute back using $ \sin x + \cos x = u $ and the trigonometric identity $ \sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + (\frac{u}{\sqrt{2}})^2} = \sqrt{\frac{2 + u^2}{2}} $ to get:

$$ = \frac{1}{\sqrt{2}} \ln|\sqrt{\frac{2 + u^2}{2}} + \frac{u}{\sqrt{2}}| + C $$

$$ = \frac{1}{\sqrt{2}} \ln|\sqrt{\frac{2 + (\sin x + \cos x)^2}{2}} + \frac{\sin x + \cos x}{\sqrt{2}}| + C $$