What is #int cos^2xsinx-tan^2xcotx dx#?

Question

Charles Arocha · Accepted Answer

#int cos^2(x)sin(x)-tan^2(x)cot(x) dx = lnabs(cos(x))-cos^3(x)/x+C#

Divide the integral in half.

#int cos^2(x)sin(x)-tan^2(x)cot(x) dx = int cos^2(x)sin(x)dx - int tan^2(x)cot(x) dx#

First Action

For #int cos^2(x)sin(x)dx#,

Let #u=cos(x)# and thus #du=-sin(x)dx#

When you swap, you obtain

#int cos^2(x)sin(x)dx=-int cos^2(x)(-sin(x)dx)=-int u^2du=-u^3/3+C =-cos^3(x)/3+C#

Step Two

For #int tan^2(x)cot(x) dx#,

Since #tan^2(x)cot(x)=tan^2(x)/tan(x)=tan(x)#

#int tan^2(x)cot(x)=int tan(x)dx#

Currently, you can obtain the answer instantly by using a formula sheet, but if you're interested, you can proceed with the following steps.

#int tan(x)dx = int sin(x)/cos(x) dx#

Now if you let #w=cos(x)#, then #dw=-sin(x)dx#

#int sin(x)/cos(x) dx = -int 1/cos(x) * -sin(x)dx=-int 1/w dw = -lnabs(w)+C=lnabs(cos(x))+C#

Last action

Thus, one obtains by deducting the two integrals

#int cos^2(x)sin(x)-tan^2(x)cot(x) dx = -cos^3(x)/x - (-lnabs(cos(x)))+C = lnabs(cos(x))-cos^3(x)/x+C#

William Abdalla · Answer

undefined To evaluate the integral $\int \cos^2(x) \sin(x) - \tan^2(x) \cot(x) \, dx$, we can use trigonometric identities to simplify the expression.

$\cos^2(x) = 1 - \sin^2(x)$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$ so $\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}$. Similarly, $\cot(x) = \frac{1}{\tan(x)}$.

Substituting these identities into the original integral, we get:

$\int \left((1 - \sin^2(x))\sin(x) - \frac{\sin^2(x)}{\cos^2(x)} \cdot \frac{1}{\tan(x)}\right) \, dx$

Now, simplify and integrate term by term:

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

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

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

Integrate each term separately:

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

Where $C$ is the constant of integration.