Question

Integrate

Answer 1

Int[dx/(sinx+secx)]

Answer 2

\[\int\limits \frac{dx}{sinx+secx}\]
That?

Answer 3

Weierstrass substitution

Answer 4

@SithsAndGiggles
you have missed square in 1-t^2 for secx part
any way this may become long ...

Answer 5

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[\int\frac{dx}{\sin x+\sec x}\]
Let \(t=\tan\left(\dfrac{x}{2}\right)\), giving \(dt=\dfrac{1}{2}\sec^2\left(\dfrac{x}{2}\right)~dx\).

Here's a table of things you'll need that are derived from the substitution:
\[\begin{matrix}\sin x=\frac{2t}{1+t^2}&&\cos x=\frac{1-t^2}{1+t^2}\\\\
\cos^2\left(\dfrac{x}{2}\right)=\frac{1}{1+t^2}\end{matrix}\]
So your integral changes to
\[{\huge\int}\frac{\dfrac{2}{1+t^2}}{\dfrac{2t}{1+t^2}+\dfrac{1+t^2}{1-t^2}}~dt\]
\(\color{blue}{\text{End of Quote}}\)

Answer 6

@SithsAndGiggles
cos x = (1 -tan(x/2)^2 ) / (1 + tan(x/2)^2 )

Answer 7

Is that better?

Answer 8

i know it was typo error

Answer 9

@shubham.bagrecha
1/(sinx + secx)
= cosx/(sinxcosx+1)
=2cosx/ (2+sin2x)
=(cosx+sinx+cosx-sinx) / (2+sin2x)
=(cosx+sinx)/ (2+sin2x) + (cosx-sinx)/ (2+sin2x)
=(cosx+sinx)/(3-(1-sin2x)) + (cosx-sinx)/ (1+1+sin2x)
=(cosx+sinx)/(3-(sinx-cosx)^2) + (cosx-sinx)/ (1+(sinx+cosx)^2)
=d(sinx-cosx) /(3-(sinx-cosx)^2) + d(sinx+cosx) / ((1+(sinx+cosx)^2)
now both the parts can be integrated using standard integral.....