Mfrhtyj

I'm having a problem with substituting when it comes to trig functions.

The integral is:$$intsqrt{1+sin(x)}dx$$

Substituting: $1+sin(x) = u implies frac{d}{dx} u = cos(x) implies dx = frac{1}{cos(x)}du$

So now we have the integral: $$int frac{sqrt{u}}{cos(x)}du$$

So now the question is, what do I do with the $frac{1}{cos(x)}$? How can I substitute that with $u$?

Hint Write your integrand in the form
$$sqrt{frac{(1+sin(x))(1-sin(x))}{1- sin(x)}}=frac{pmcos(x)}{sqrt{1-sin(x)}}$$
and substitute $$t=1-sin(x)$$ then we get $$dt=-cos(x)dx$$

HINT

Here I give you an alternative approach. Observer that

begin{align*}
1 + sin(x) & = 1 + 2sinleft(frac{x}{2}right)cosleft(frac{x}{2}right) = left[cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right]^{2}
end{align*}

Hence the given integral is equal to

begin{align*}
intsqrt{1+sin(x)}mathrm{d}x = intleft|cosleft(frac{x}{2}right) + sinleft(frac{x}{2}right)right|mathrm{d}x
end{align*}

Hint:
$$
begin{align}
sqrt{frac{1+sin(x)}2}
&=sqrt{frac{1+cosleft(fracpi2-xright)}2}\
&=left|,cosleft(tfracpi4-tfrac x2right),right|
end{align}
$$