Integral: Solve $\int \frac{\sin x+\cos x}{\sec x+ \tan x}dx$

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Homework Statement


\int \frac{\sin x+\cos x}{\sec x+ \tan x}dx

Homework Equations



\sin x = \frac{1}{\sec x}
\cos x = \frac{\sin x}{\tan x}


The Attempt at a Solution


i separate and try to use identities but i got nothign

1/(secx^2+secx tan x)+sin x/tanx^2+sec x :confused:
 
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write everything in terms of sin x and cos x only, then integrate with change of variable.
 
One of your identities is incorrect: sec(x)=1/cos(x).

As has been said above, you should first look to express everything in terms of sines and cosines. See if this gives you a hint as to how to proceed.
 
please tell more

this is the shape that i got
don't know how to complete

–integral sin²x-sin2x-1/2(1+sinx)
 
-(\int \sin^2 x dx - \int \sin 2x dx - 1/2\int 1+\sin x)

Write sin^2 x as (1/2) (1-cos2x).

for sin 2x, make a substitution u=2x, then remember the integral of sin u is -cos u.

For the 3rd one, if you can't do it, why are you doing this question?
 
you can also try weierstrass substitution
i.e t = tan x/2
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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