Question

integrate1/cosx+sinx

Answer 1

Do you mean \[\int\limits_{}^{} 1 / (\sin x + \cos x)\]

Answer 2

\[1-2\sin ^{2}x=u , 2\cos ^{2}x-1=v\]
\[du=-2(2)cosdx=-4cosdx, dv=-4sindx\]
substitute in the equation.

Answer 3

do you think it works??

Answer 4

yes it should work....
u will get cosx=-1/4 du and sinx=-1/4dv

Answer 5

so what??? @aryandecoolest ? are you kidding?

Answer 6

so 1/4 (du)/u - Int (-1/4) dv/v
which gives us 1/4 ln (2cos2x-1)-1/4 ln (1-2sin2x) +C

Answer 7

:?????????????????????????????

Answer 8

Int dx/(sinx+cosx)= Int (cosx-Sinx)dx/(Cos2x-sin2x)= Int cosx dx/ (1-2Sin2x) - Int sinx dx/ (2cos2x-1) ..(.1)

Let 1-2sin2x =u and 2cos2x-1 =v

so, du= -2 (2)cosx dx =-4cosx dx, thus cosx dx = -1/4 du........(2)

and dv= -2 (2) sinx dx = -4sin x dx, thus sinx dx= -1/4 dv....(3)

let's now plug-in the value of sin x and cosx in (1)

We have int -1/4 (du)/u - Int (-1/4) dv/v

=-1/4 ln(u) -1/4 ln (v)+C

= -1/4 ln (2cos2x-1)-1/4 ln (1-2sin2x) +C

Answer 9

This can be done with the half-angle substitution: \(t=\tan\dfrac{x}{2}\). This gives you a list of facts:
\[\sin x=\sin\left(2\cdot\frac{x}{2}\right)=2\sin\frac{x}{2}\cos\frac{x}{2}\\
\cos x=\cos\left(2\cdot\frac{x}{2}\right)=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\\
dt=\frac{1}{2}\sec^2\frac{x}{2}~dx~~\iff~~2\cos^2\frac{x}{2}~dt=dx\]
Given that \(t=\tan\dfrac{x}{2}\), you can set up a triangle:

so that
\[\sin x=\frac{2t}{1+t^2}\\
\cos x=\frac{1-t^2}{1+t^2}\\
\frac{2}{1+t^2}~dt=dx\]
We then equate the integrals under this substitution:
\[\int\frac{dx}{\cos x+\sin x}={\Huge\int}\frac{\dfrac{2}{1+t^2}}{\dfrac{2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}}~dt\]
Simplifying gives
\[\int\frac{2}{2t+1-t^2}~dt\]
Completing the square in the denominator gives
\[\int\frac{2}{2-(t-1)^2}~dt\]
which can be taken care of with a trig sub: \(t-1=\sqrt2\sin u\), so that \(dt=\sqrt 2\cos u~du\).
\[\int\frac{2\sqrt2\cos u}{2-(\sqrt2\sin u)^2}~du\\
\sqrt2\int\frac{\cos u}{1-\sin^2 u}~du\\
\sqrt2\int\frac{\cos u}{\cos^2 u}~du\\
\sqrt2\int\frac{du}{\cos u}\\
\sqrt2\int\sec u~du\]