How to calculate $\int \frac{dx}{\cos(x) + \sin(x)} $?

Question

I did it by the method of integration by parts, with

$$ u=\frac{dx}{\cos(x) + \sin(x)},\quad dv=dx$$

so $$ \int \frac{dx}{\cos(x) + \sin(x)} = \frac{x}{\cos(x)+\sin(x)} - \int \frac{x(\sin(x)-\cos(x))}{(\cos(x) + \sin(x))^{2}} $$ Where,
$$\int \frac{x(\sin(x)-\cos(x))}{(\cos(x) + \sin(x))^{2}}\;dx = \int \frac{x \sin(x)}{1+2\sin(x)\cos(x)}dx - \int \frac{x \cos(x)}{1+2\sin(x)\cos(x)}dx , $$ I have not managed to solve those two integrals that were expressed, really appreciate if you can help me.

Answer 1

$$\int \frac{dx}{\cos(x) + \sin(x)} $$ $$=\int \frac{dx}{\sqrt2\left(\sin (x)\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\cos(x) \right)} $$ $$=\int \frac{dx}{\sqrt2\sin \left(x+\frac{\pi}{4} \right)} $$ $$=\frac1{\sqrt2}\int \csc \left(x+\frac{\pi}{4} \right)\ d\left(x+\frac{\pi}{4} \right) $$ Using formula: $\int\csc\theta d\theta=\ln\left|\tan\left(\frac{\theta}{2}\right)\right|$, $$=\frac1{\sqrt2}\ln\left|\tan\left(\frac{x+\frac{\pi}{4}}{2}\right)\right|+C$$ $$=\frac1{\sqrt2}\ln\left|\tan\left(\frac x2+\frac{\pi}{8}\right)\right|+C$$

Answer 2

Use$$\int\frac{dx}{\cos x+\sin x}=\int\frac{1}{\sqrt{2}}\sec(x-\pi/4)dx=\frac{1}{\sqrt{2}}\ln|\sec(x-\pi/4)+\tan(x-\pi/4)|+C.$$

Answer 3

hint

Make the substitution $$t=\tan(\frac x2)$$with $$\sin(x)=\frac{2t}{1+t^2}\;\;,\;\cos(x)=\frac{1-t^2}{1+t^2}\;\;dx=\frac{2dt}{1+t^2}$$

the integral becomes $$-2\int \frac{dt}{t^2-2t-1}=$$ $$\frac 12\int (\frac{\sqrt{2}}{(t-1+\sqrt{2}}-\frac{\sqrt{2}}{t-1-\sqrt{2}})dt$$

Answer 4

There are a number of ways to compute the antiderivative, but integration by parts is not one of them.

One method that was used was to recall the angle addition identity $$\cos (x - \theta) = \cos x \cos \theta + \sin x \sin \theta,$$ and letting $\theta = \pi/4$ gives $$\cos \left(x - \frac{\pi}{4}\right) = \frac{\cos x}{\sqrt{2}} + \frac{\sin x}{\sqrt{2}} = \frac{\cos x + \sin x}{\sqrt{2}},$$ from which it follows that $$\frac{1}{\sin x + \cos x} = \sqrt{2} \sec \left( x - \frac{\pi}{4}\right)$$ as claimed.

Answer 5

$$\int \frac{dx}{\cos(x) + \sin(x)}=\int \frac{\cos(x)-\sin(x)}{\cos^2(x) -\sin^2(x)}dx$$

$$=\int \frac{\cos(x)}{\cos^2(x) - \sin^2(x)}dx-\int \frac{\sin(x)-\sin(x)}{\cos^2(x) - \sin^2(x)}dx$$

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

$$=\frac{1}{\sqrt{2}}\tanh^{-1}\left(\sqrt{2}\sin(x)\right)-\frac{1}{\sqrt{2}}\tanh^{-1}\left(\sqrt{2}\cos(x)\right)$$

Answer 6

$\int\frac{1}{\sin(x)+\cos(x)}dx=\int\frac{1} {a\sin(x)+b\cos(x)}dx$
where $a=b=1$
We put $ a=r\sin(y), b=r\cos(y) ,r=\sqrt{a^2+b^2}, y=\tan^{-1}(\frac{a}{b})$
\begin{align*} \int\frac{1}{a\sin(x)+b\cos(x)}dx=\int\frac{dx}{r\sin(y)\sin(x)+r\cos(y)\cos(x)}\\ &=\frac{1}{\sqrt{a^2+b^2}}\int\frac{dx}{\cos(x+y)}\\ &=\frac{1}{\sqrt{a^2+b^2}}\int\frac{\sec(x+y)(\sec(x+y)+\tan(x+y))}{\sec(x+y)+\tan(x+y)}dx\\ &=\frac{1}{\sqrt{a^2+b^2}}\log(\sec(x+y)+\tan(x+y))\\ &=\frac{1}{\sqrt{a^2+b^2}}\log(\frac{1}{\cos(x+\tan^{-1}(\frac{a}{b})}+\frac{\sin(x+\tan^{-1}(\frac{a}{b}))}{\cos(x+\tan^{-1})(\frac{a}{b}))})\\ &=\frac{1}{\sqrt{a^2+b^2}}\log(\frac{\sqrt{a^2+b^2}+\sqrt{a^2+b^2}\sin(x+\tan^{-1}(\frac{a}{b}))}{a\sin(x)+b\cos(x)}) \end{align*} we put $ a=1, b=1$
$\int\frac{1}{\sin(x)+\cos(x)}dx=\frac{1}{\sqrt{2}}\log(\frac{\sqrt{2}+\sqrt{2}\sin(x+\tan^{-1}(1))}{\sin(x)+\cos(x)})$