# trig function

• Oct 22nd 2007, 10:59 AM
bobak
trig function
Hello all

here is my problem

$\displaystyle \int {\sqrt{1+\sin x}\,dx}$

i tired a half angle approach ( and didnt get very far). any hints?
• Oct 22nd 2007, 11:29 AM
ticbol
Quote:

Originally Posted by bobak
Hello all

here is my problem

$\displaystyle \int {\sqrt{1+\sin x}\,dx}$

i tired a half angle approach ( and didnt get very far). any hints?

Hint only?

cos^2(X) = 1 -sin^2(X)
so,
cosX = sqrt[1 -sin^2(X)]
cosX = sqrt[(1 +sinX)(1 -sinX)]

Further,
INT.[sqrt(1 +sinX)]dX
= INT.[sqrt(1 +sinX)]*[cosX / cosX] dX
• Oct 22nd 2007, 11:45 AM
bobak
$\displaystyle -2\sqrt{1-\sin x}$ that right ?
• Oct 22nd 2007, 12:07 PM
ticbol
Yes.

But you have to add the all-impotant "C".

= -2sqrt(1 -sinX) +C
• Oct 22nd 2007, 02:36 PM
Soroban
Hello,bobak!

Quote:

Here is my problem: .$\displaystyle \int \sqrt{1+\sin x}\,dx$
Multiply by $\displaystyle \frac{1-\sin x}{1-\sin x}$

. . $\displaystyle \sqrt{\frac{1-\sin x}{1-\sin x}\cdot(1+\sin x)}\;=\;\sqrt{\frac{1-\sin^2x}{1-\sin x}} \;=\;\sqrt{\frac{\cos^2x}{1-\sin x}}\;=\;\frac{\cos x}{\sqrt{1-\sin x}}$

We have: .$\displaystyle \int\frac{\cos x}{\sqrt{1-\sin x}}\,dx \;=\;\int(1-\sin x)^{-\frac{1}{2}}(\cos x\,dx)$

Now let: $\displaystyle u \,= \,1-\sin x$