What is #int sin^3x/cos^6x dx#?

Question

What is #int sin^3x/cos^6x  dx#?

Isaiah Allen · Accepted Answer

#int frac{sin^3x}{cos^6x} dx = -frac{1}{3cos^3x} + frac{1}{5cos^5x} + c#,

where #c# is the constant of integration.

Use the substitution #u=cosx#. #frac{du}{dx} = -sinx# #int frac{sin^3x}{cos^6x} dx = int frac{cos^2x-1}{cos^6x} (-sinx) dx# #= int frac{u^2-1}{u^6} frac{du}{dx} dx# #= int (u^{-4} - u^{-6}) du# #= frac{u^{-3}}{-3} - frac{u^{-5}}{-5} + c#, where #c# is the constant of integration. #= -frac{1}{3cos^3x} + frac{1}{5cos^5x} + c#

Paisley Johnson · Answer

undefined To integrate $\frac{{\sin^3(x)}}{{\cos^6(x)}}$ with respect to $x$, use the substitution $u = \cos(x)$. This will change the integral into one involving only $u$. Then use trigonometric identities to simplify the resulting integral. The steps are as follows:

1. Let $u = \cos(x)$, then $du = -\sin(x)dx$.
2. Substitute $-\sin(x)dx = du$ and $\sin^2(x) = 1 - \cos^2(x)$ into the integral.
3. Rewrite the integral entirely in terms of $u$.
4. Simplify the integral using trigonometric identities.
5. Integrate the resulting expression with respect to $u$.
6. Finally, resubstitute $u = \cos(x)$ back into the expression.

The final result is:

$$
\int \frac{{\sin^3(x)}}{{\cos^6(x)}} dx = \frac{{-\sin^2(x)}}{5\cos^5(x)} + C
$$

where $C$ is the constant of integration.

Paisley Johnson · Answer

undefined To integrate $ \frac{{\sin^3(x)}}{{\cos^6(x)}} $ with respect to $ x $, you can use the trigonometric identity $ \sin^2(x) = 1 - \cos^2(x) $ to rewrite the integral in terms of $ \cos(x) $. Then make a substitution to simplify the integral.

Here's the solution:

Let $ u = \cos(x) $, so $ du = -\sin(x) \, dx $.

Now substitute $ u $ and $ du $ into the integral:

$$ \int \frac{{\sin^3(x)}}{{\cos^6(x)}} \, dx = -\int \frac{{\sin^2(x) \sin(x)}}{{\cos^6(x)}} \, dx $$

Using the trigonometric identity $ \sin^2(x) = 1 - \cos^2(x) $:

$$ = -\int \frac{{(1 - \cos^2(x)) \sin(x)}}{{\cos^6(x)}} \, dx $$

$$ = -\int \frac{{(1 - u^2)(-du)}}{{u^6}} $$

$$ = \int \frac{{(1 - u^2)}}{{u^6}} \, du $$

$$ = \int \left( u^{-6} - u^{-4} \right) \, du $$

$$ = \frac{{u^{-5}}}{{-5}} - \frac{{u^{-3}}}{{-3}} + C $$

$$ = -\frac{1}{5u^5} + \frac{1}{3u^3} + C $$

Finally, substitute $ u = \cos(x) $ back:

$$ = -\frac{1}{5\cos^5(x)} + \frac{1}{3\cos^3(x)} + C $$

So, the integral of $ \frac{{\sin^3(x)}}{{\cos^6(x)}} $ with respect to $ x $ is $ -\frac{1}{5\cos^5(x)} + \frac{1}{3\cos^3(x)} + C $, where $ C $ is the constant of integration.