Question

Calculus 2 master needed; evaluate the integral PLEASE SHOW STEPS IF IM WRONG x%2F%5Csqrt%7Bcosx%7D%20%7D%20%5C%2C%20dx" id="TexFormula1" title="\int{sin^3x/\sqrt{cosx} } \, dx" alt="\int{sin^3x/\sqrt{cosx} } \, dx" align="absmiddle" class="latex-formula"> I split off the sin^3 so i can use the pythag identity and allows for u substitution u=cosx du=-sinx dx -du=sin dx $\int{1-u^2/\sqrt{u}*-du }$ I move the negative towards the outside of the integral. then i divide the terms by sqroot 2||| $-\int{(1/\sqrt{u} - u^2/\sqrt{u} )} \, du$ I eventually get to a=1/2 b =5/2 $-2{cos^a x +2/5cos^b} \, dx$ did I miss anything? Or is this the final answer?

Answer 1

Answer:

Yes, you answer is correct! It just needs to be simplified :)

Step-by-step explanation:

So we have the integral:

$\int \frac{\sin^3(x)}{\sqrt{\cos(x)}}dx$

As you had done, we can split off the numerator:

$=\int \frac{\sin(x)(\sin^2(x))}{\sqrt{\cos(x)}}dx$

Using the Pythagorean Identity, this is:

$=\int \frac{\sin(x)(1-\cos^2(x))}{\sqrt{\cos(x)}}dx$

Now, we can do u-substitution. Let u equal cos(x). Thus:

$u=\cos(x)\\du=-\sin(x)dx\\-du=\sin(x)dx$

So:

$=\int \frac{1-u^2}{\sqrt{u}}(-du)$

Simplify:

$=-\int\frac{1-u^2}{\sqrt u}du$

We can then split the terms:

$=-\int \frac{1}{\sqrt u}-\frac{u^2}{\sqrt u}du$

Expand the integral:

$=-(\int \frac{1}{\sqrt u}du-\int\frac{u^2}{\sqrt u}du)$

Simplify each of the u.

For the left, that is simply u^-1/2.

For the right, it is u^(2-1/2) or u^3/2. Thus:

$=-(\int u^{-\frac{1}{2}}du-\int u^{\frac{3}{2}}du)$

Reverse Power Rule:

$=-(\frac{u^{1+-\frac{1}{2}}}{1+-\frac{1}{2}}-\frac{u^{1+\frac{3}{2}}}{1+\frac{3}{2}})$

Simplify:

$=-(\frac{u^{\frac{1}{2}}}{\frac{1}{2}}-\frac{u^{\frac{5}{2}}}{\frac{5}{2}})$

Simplify further:

$=-(2u^{\frac{1}{2}}-\frac{2u^{\frac{5}{2}}}{5})$

Distribute the negative:

$=-2u^{\frac{1}{2}}+\frac{2u^{\frac{5}{2}}}{5}$

And substitute back cos(x) for u:

$=-2\cos^{\frac{1}{2}}(x)+\frac{2\cos^{\frac{5}{2}}(x)}{5}$

And this is precisely what you got, so well done!

We can simplify this by first multiplying the first term by 5 to get a common denominator. So:

$=-\frac{10\cos^{\frac{1}{2}}(x)}{5}+\frac{2\cos^{\frac{5}{2}}(x)}{5}$

Combine:

$=\frac{-10\cos^{\frac{1}{2}}(x)+2\cos^{\frac{5}{2}}(x)}{5}$

Factor out a cos^(1/2)(x) and a 2. Since we factored out a cos^(1/2)(x), we need to subtract their exponents inside. Thus:

$=\frac{2\cos^{\frac{1}{2}}(x)(-5\cos^{\frac{1}{2}-\frac{1}{2}}(x)+\cos^{\frac{5}{2}-\frac{1}{2}}(x))}{5}$

Simplify:

$=\frac{2\cos^{\frac{1}{2}}(x)(-5+\cos^2(x))}{5}$

Simplify:

$=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}$

And, of course, C:

$=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}+C$

So:

$\int \frac{\sin^3(x)}{\sqrt{\cos(x)}}dx=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}+C$

And we're done :)

Answer 2

Answer:

=  - 2 $\sqrt{cos(x)}$ + 2 cos⁵/₂ (x)  + C

                          5

Step-by-step explanation:

∫ sin³ (x)    dx

  $\sqrt{cos(x)}$

= ∫ sin² (x) sin (x)    dx

   $\sqrt{cos(x)}$

= ∫ (1 - cos² (x) sin (x)  dx

   $\sqrt{cos(x)}$

= ∫ - 1 - u²   du

        √u

= ∫ -   1    +  u³/₂   du

        √u

= - ∫   1    du  +  ∫ u³/₂   du

        √u

substitute it back

=  - 2 √u + 2 cos⁵/₂ (x)

                 5

add constant, therefore

=  - 2 $\sqrt{cos(x)}$ + 2 cos⁵/₂ (x)  + C

                         5