Question

∫1/(sinx+cosx)dx

Answer 1

write the denominator as
(1/sqrt 2 sin x + 1/sqrt 2 cos x)
so that we get the form of sin A cos B + cos A sin B

Answer 2

you know what sin A cos B + cos A sin B =...?

Answer 3

see whether you get the denominator as
sin (x + pi/4)
then it'd be easy to integrate

Answer 4

Weierstrass substitution works best for these types of functions. Let \(t=\tan\left(\dfrac{x}{2}\right)\).

Answer 5

So if \(t=\tan\left(\dfrac{x}{2}\right)\), you have \(dt=\dfrac{1}{2}\sec^2\left(\dfrac{x}{2}\right)~dx\).

From the intitial substitution, you know that (drawing a triangle would help you see why)
\[\begin{align*}\cos\left(\dfrac{x}{2}\right)&=\frac{1}{\sqrt{1+t^2}}\\
\sin\left(\dfrac{x}{2}\right)&=\frac{t}{\sqrt{1+t^2}}
\end{align*}\]
Using the double angle identities, you also know that
\[\cos x=\cos2\left(\dfrac{x}{2}\right)=\cos^2\left(\dfrac{x}{2}\right)-\sin^2\left(\dfrac{x}{2}\right)\\
\sin x=\sin2\left(\dfrac{x}{2}\right)=2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)
\]
From the above, you then know that
\[\cos x=\frac{1}{1+t^2}-\frac{t^2}{1+t^2}=\frac{1-t^2}{1+t^2}\\
\sin x=2\frac{t}{\sqrt{1+t^2}}\frac{1}{\sqrt{1+t^2}}=\frac{2t}{1+t^2}
\]
and that
\[\begin{align*}dt=\frac{1}{2}\sec^2\left(\dfrac{x}{2}\right)~dt~~\Rightarrow~~2\cos^2\left(\frac{x}{2}\right)~dt&=dx\\
\frac{2}{1+t^2}~dt&=dx
\end{align*}\]

Answer 6

So, finally, and assuming I haven't made any mistakes above, you get
\[\int\frac{dx}{\sin x+\cos x}=\int\frac{\frac{2}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}~dt=\int\frac{2}{1+2t-t^2}~dt\]

Answer 7

then ? :P

Answer 8

Completing the square would work, along with a trigonometric substitution.

Answer 9

whoa!
yes, that would surely work.
did you try it my way ? without Weierstrass substitution
my point is to compare the complexity of the solution :)